There are a few ways in calculus to find a max or a min. These ways are listed below:
You can: take the derivative and set equal to zero and solve to find the extrema of the given function then use the x's you got when you solved to plug into the original equation and solve to get the largest result as the absolute max and the smallest result as the absolute min Example: using f(X)= 3x^4-4x^3 on the interval [-1,2] determine the aboslute max or min
take the derivative and set equal to zero and solve to find the extrema of the given function
4(3)x^3-3(4)x^2 12x^3-12x^2=0 12x^2(x-1)=0 x= 0,1
then use the x's you got when you solved and the given endpoints of the interval to plug into the original equation and solve
therefore f(2)3(2)^4-4(2)^3=16 is the absolute max
the smallest result is the absolute min
therefore f(1)3(1)^4-4(1)^3= -1 is the absolute min
You can also use the first derivative test to find a max or min
Steps: similar to the last method of getting the max or min I mentioned Take derivative and set equal to zero
Solve for x
Set up intervals using(-infinity, pt)U(pt, infinity)
Example:Plug in value on the interval into the derivative ((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))ans then to find a max or a min remember(( A max is increasing then decreasing, A min is decreasing then increasing))
Find the open intervals on which f(x)= x^3-3/2x^2 is increasing or decreasing Take derivative and set equal to zero 3x^2-3x=0 Solve for x 3x(x-1)=0 X=0,1 Set up intervals (-infinity, pt)U(pt, infinity) (negative infinity,0)U(0,1)U(1,infinity) Plug in value on the interval into the derivative F’(-1)= 3(-1)^2-3(-1)=6 Positive therefore increasing F’(1/2)= 3(1/2)^2-3(1/2)= -3/4 Negative therefore decreasing F’(2)= 3(2)^2-3(2)= 6 Positive therefore increasing Therefore increasing: (negative infinity, 0) U (1, infinity) Decreasing: (0,1) Max: at 0 because at 0 the function is changing from increasing to decreasing Min: at 1 because at 1 the function is changing from decreasing to increasing.
Also, you can use a picture of a graph by looking for the highest point (max)and the lowest point(min)
or the calculator to find the max or min by looking for the highest and lowest points in the graph respectively. When using the calculator you will type the given function into y equals and press graph to view the image of the graphed function and use the trace function paired with either the max or min function depending on whether you want to find the highest point which will be the max or the lowest point which will be the min
There is also the short cut to the first derivative test that will get you the max or min.
Steps to the shortcut for the first derivative test:
Step #1: take the derivative, set equal to zero, and solve
Step #2: Check to see if differentiable
Step #3: Plug in critical points into the second derivative
Step #4 If POSITIVE there is a min If NEGATIVE there is a max If 0 the test is false (((NOTICE THAT FOR THE SHORTCUT WHETHER THE PLUG IN IS POSITIVE OR NEGATIVE HAS A REVERSED MEANING))) Example:
Step #1: take the derivative, set equal to zero, and solve -15x^4+15x^2=0 15x^2(-x^2+1) x= 0, 1,-1
Step #2: Check to see if differentiable differentiable? yes.
Step #3: Plug in critical points into the second derivative F"(X)=60x^3 + 30x -60(0)^3+30(0)=0 -60(-1)^3+30(-1)=POSITIVE -60(1)^3+30(1)=NEGATIVE
Step #4 If POSITIVE there is a min Therefore, at x= -1 there is a min If NEGATIVE there is a max Therefore, at x= 1 there is a max If 0 the test is false Therefore, at x=0 the test fails
I have already mentioned how to tell is the value is a max or a min but i will reiterotiate((sp?)
Determining whether the value is a max or a min is depenent upon the method you chose to use to find the max's and min's.
Using the extrema/interval method: when you solved to plug into the original equation and solve to get the largest result as the absolute max and the smallest result as the absolute min
Using the first derivative test: Plug in value on the interval into the derivative ((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))ans then to find a max or a min remember(( A max is increasing then decreasing, A min is decreasing then increasing))
Using a picture of a graph: by looking for the highest point (max)and the lowest point(min)
using a calculator: use the trace function paired with either the max or min function depending on whether you want to find the highest point which will be the max or the lowest point which will be the min
Using the shortcut to the first derivative test: Plug in critical points into the second derivative If POSITIVE there is a min If NEGATIVE there is a max (((AGAIN, NOTICE THAT FOR THE SHORTCUT WHETHER THE PLUG IN IS POSITIVE OR NEGATIVE HAS A REVERSED MEANING)))
Using Calculus how can you find a max or a min? -There are many different ways to find a max or min in Calculus. All of the one’s we learned so far were in chapter three. To find max or min you have to take the derivative, just like everything else in Calculus. One way is using the first derivative test.
Steps for the first derivative test: tep 1: Take the derivative. Step 2: Set it equal to zero. Step 3: Solve for x. Step 4: Set up intervals. (-infinity, _ ) u ( _, infinity) Step 5: Plug a value on the interval into the derivative. If you get a positive number, the function is increasing. If you get a negative number, the function is decreasing. Step 6: Figure out the max and min. (A max is increasing then decreasing and a min is decreasing then increasing.)
Another way is finding the Critical number, etc. is the same as finding a horizontal tangent. These will also give you a max and min. To find the max and min, you take the derivative, then set equal to zero. If you have an interval and it asks for the max and min, plug your x-values you get from solving the derivative and your endpoints from the interval into the original equation. The biggest number is your max and the smallest number is your min.
Ex. Find the max and min on the graph f(x) = cos π x on the interval of [0, 1/6]. f ’(x) = (-sin π x) (π) = -π sin π x -π sin π x = 0 sin π x = 0 π x = sin^-1(0) π x = 0, π, 2π x = 0, 1, 2 f(0) = cos π (0) = cos 0 = 1 f(1/6) = cos π (1/6) = cos π/6 = ½ f(1) = cos π (1) = cos π = -1 f(2) = cos π (2) = cos 2π = 1 max = 1 min = -1
**HELPFUL WORDS: Extreme values – max and min Absolute max and min – the highest and lowest point on an interval Relative max and min – any max or min not on an interval Critical number – any max or min (often referred to as x = c)
One way to find a max/min using Calculus is the First Derivative Test:
Step 1: Take the derivative. Step 2: Set it equal to zero. Step 3: Solve for x. Step 4: Set up intervals. (-infinity, _ ) u ( _, infinity) Step 5: Plug a value on the interval into the derivative. If you get a positive number, the function is increasing. If you get a negative number, the function is decreasing. Step 6: Figure out the max and min. (A max is increasing then decreasing, and a min is decreasing then increasing.)
Ex. Find the max and min of f(x) = x^3 – 3/2x^2. 1) 3x^2 – 3x = 0 2) 3x (x – 1) = 0 3) x = 0, 1 4) (-infinity, 0) u (0, 1) u (1, infinity) 5) f’(-1) = 3(-1)^2 – 3(-1) = 6 increasing f’(1/2) = 3(1/2)^2 – 3(1/2) = -3/4 decreasing f’(2) = 3(2)^2 – 3(2) = 6 increasing 6) x = 0 is a max & x = 1 is a min
*Note: If they ask for absolute extrema, you have to plug back into the original equation to find the y-values in order to find the critical points and end points.
**Soooo, how can you find a max or a min? Hmm..first let's refresh our memories on section 3.2, shall we? Well in that section we were introduced to Rolle's Theorem. By applying that theorem, it tells you IF there is a max or min on an interval, and it guarantees that you have AT LEAST 1 max or min..Let's not get that confused with the First Derivative Test though. The First Derivative Test is where you actually can find the max's/mins. -To do that, you first take the derivative of the function. -Then you set it equal to zero and solve for x (remember, if it happens to be a fraction, you just set the numerator equal to zero and solve) -Once you get your x-value(s), you check for differentiability. -Then you set up intervals starting with negative infinity and ending with infinity -Then you plug values within those intervals into the 1st derivative..to see if you get a positive or negative value (if positive= INCREASING; if negative=DECREASING) **How to determine if it's a max/min.. -If you have a change going from decreasing to increasing, (neg. to pos.) that means you have a min at whatever x-value it is. -If you have a change going from increasing to decreasing (pos. to neg.) that means you have a max at whatever the x-value is. -If you happen to get increasing followed by increasing, then it's nothing..there are no max's or mins. That also goes for if you have decreasing followed by decreasing..
1. Take the derivative 2. Set the derivative = 0 and solve for x. Check for differentiability after 3. Set up your intervals from -infinity to infinity 4. Plug in values from your interval. Positive results mean it is increasing and negative results are decreasing 5. If the value changes from decreasing to increasing, it is a min on that point. If the value changes from increasing to decreasing it is.... you guessed it! a max on that point.
To find a max or min using calculus, you just use the first derivative test. first, you take the first derivative. second, you set the derivative equal to zero third, you solve for x. fourth, you set up intervals (-infinity, x)u(x, infinity) then, you plug numbers from each interval into the first derivative. if its positive, it is increasing. if it is negative, then it is decreasing. if the function is increasing, then decreasing, you have a max. if it is decreasing then increasing, you have a min. finally, you state the x-intercept that is your max or min (x=# is a max/min by the first derivative test).
Using Calculus how can you find a max or a min? How do you determine if the value is a max or min?
To find a max or a min of a function using calculus you must use the first derivative test. to use the first derivative test you must first take the derivative of the equation of the function. The you must solve the derivative for x. the answers you get are your critical points. your critical points are the maxs or mins of the function. to find whether a critical point is a max or a min then you must set up the intervals the function using the critical points. plug in a number within the interval into the derivative of the equation. if you get a positive number then that critical point is a max, if you get a negative then that critical point is a min.
There are different ways to find a max or min in Calculus.
One simple way to find maxes or mins is the first derivative test.
1. Find first derivative. 2. Set it equal to zero. 3. Solve for x. 4. Set up intervals for which necessary. 5. Plug in numbers between each number on each interval to the first derivative. 6. If the outcome is positive, it is increasing. If the outcome is negative, it is decreasing. 7. Finally you have to find where it increasing to decreasing, or decreasing to increasing. If it is increasing to decreasing, the point of that is said to be a max. If it is decreasing, the point of that is said to be a min.
Finding a max or min in calculus is pretty simple. All you have to do is use the 1st derivative test. Steps for the first derivative test are:
1. Take the first derivative 2. Set it equal to zero and solve for x. 3. Check to see if it is differentiable. 4. Set up intervals for the numbers you found from steps 2 and 3. 5. Plug in numbers on each interval. 6. If the solutions from step 5 come out positive then it is increasing, if negative then it is decreasing. 7. Maxs are at points where the interval switches from increasing to decreasing, and mins are at the points where it goes from decreasing to increasing.
There are more than one ways to find a max or min in Calculus. To find max or min you have to take the derivative, just like everything else in Calculus. Im gonna show you the first derivative test.
Steps for the first derivative test: 1: Take the derivative. 2: Set it equal to zero. 3: Solve for x. 4: Set up intervals. (-infinity, _ ) u ( _, infinity) 5: Plug a value on the interval into the derivative. If you get a positive number, the function is increasing. If you get a negative number, the function is decreasing. 6: Figure out the max and min. (A max is increasing then decreasing and a min is decreasing then increasing.)
Another way is finding the Critical number or something else. These will also give you a max and min. To find the max and min, you take the derivative, then set equal to zero. If you have an interval and it asks for the max and min, plug your x-values you get from solving the derivative and your endpoints from the interval into the original equation. The biggest number is your max and the smallest number is your min.
There are a few ways in calculus to find a max or a min. These ways are listed below:
ReplyDeleteYou can:
take the derivative and set equal to zero and solve to find the extrema of the given function then use the x's you got when you solved to plug into the original equation and solve to get the largest result as the absolute max and the smallest result as the absolute min
Example:
using f(X)= 3x^4-4x^3 on the interval [-1,2] determine the aboslute max or min
take the derivative and set equal to zero and solve to find the extrema of the given function
4(3)x^3-3(4)x^2
12x^3-12x^2=0
12x^2(x-1)=0
x= 0,1
then use the x's you got when you solved and the given endpoints of the interval to plug into the original equation and solve
f(-1)3(-1)^4-4(-1)^3=7
f(0)3(0)^4-4(0)^3=0
f(1)3(1)^4-4(1)^3= -1
f(2)3(2)^4-4(2)^3=16
the largest result is the absolute max
therefore f(2)3(2)^4-4(2)^3=16 is the absolute max
the smallest result is the absolute min
therefore f(1)3(1)^4-4(1)^3= -1 is the absolute min
You can also use the first derivative test to find a max or min
Steps:
similar to the last method of getting the max or min I mentioned Take derivative and set equal to zero
Solve for x
Set up intervals using(-infinity, pt)U(pt, infinity)
Example:Plug in value on the interval into the derivative ((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))ans then to find a max or a min remember(( A max is increasing then decreasing, A min is decreasing then increasing))
Find the open intervals on which f(x)= x^3-3/2x^2 is increasing or decreasing
Take derivative and set equal to zero
3x^2-3x=0
Solve for x
3x(x-1)=0
X=0,1
Set up intervals (-infinity, pt)U(pt, infinity)
(negative infinity,0)U(0,1)U(1,infinity)
Plug in value on the interval into the derivative
F’(-1)= 3(-1)^2-3(-1)=6
Positive therefore increasing
F’(1/2)= 3(1/2)^2-3(1/2)= -3/4
Negative therefore decreasing
F’(2)= 3(2)^2-3(2)= 6
Positive therefore increasing
Therefore increasing: (negative infinity, 0) U (1, infinity)
Decreasing: (0,1)
Max: at 0 because at 0 the function is changing from increasing to decreasing
Min: at 1 because at 1 the function is changing from decreasing to increasing.
Also, you can use a picture of a graph by looking for the highest point (max)and the lowest point(min)
or the calculator to find the max or min by looking for the highest and lowest points in the graph respectively. When using the calculator you will type the given function into y equals and press graph to view the image of the graphed function and use the trace function paired with either the max or min function depending on whether you want to find the highest point which will be the max or the lowest point which will be the min
There is also the short cut to the first derivative test that will get you the max or min.
Steps to the shortcut for the first derivative test:
Step #1:
take the derivative, set equal to zero, and solve
Step #2:
Check to see if differentiable
Step #3:
Plug in critical points into the second derivative
Step #4
If POSITIVE there is a min
If NEGATIVE there is a max
If 0 the test is false
(((NOTICE THAT FOR THE SHORTCUT WHETHER THE PLUG IN IS POSITIVE OR NEGATIVE HAS A REVERSED MEANING)))
Example:
Step #1:
take the derivative, set equal to zero, and solve
-15x^4+15x^2=0
15x^2(-x^2+1)
x= 0, 1,-1
Step #2:
Check to see if differentiable
differentiable?
yes.
Step #3:
Plug in critical points into the second derivative
F"(X)=60x^3 + 30x
-60(0)^3+30(0)=0
-60(-1)^3+30(-1)=POSITIVE
-60(1)^3+30(1)=NEGATIVE
Step #4
If POSITIVE there is a min
Therefore, at x= -1 there is a min
If NEGATIVE there is a max
Therefore, at x= 1 there is a max
If 0 the test is false
Therefore, at x=0 the test fails
I have already mentioned how to tell is the value is a max or a min but i will reiterotiate((sp?)
ReplyDeleteDetermining whether the value is a max or a min is depenent upon the method you chose to use to find the max's and min's.
Using the extrema/interval method:
when you solved to plug into the original equation and solve to get the largest result as the absolute max and the smallest result as the absolute min
Using the first derivative test:
Plug in value on the interval into the derivative ((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))ans then to find a max or a min remember(( A max is increasing then decreasing, A min is decreasing then increasing))
Using a picture of a graph:
by looking for the highest point (max)and the lowest point(min)
using a calculator:
use the trace function paired with either the max or min function depending on whether you want to find the highest point which will be the max or the lowest point which will be the min
Using the shortcut to the first derivative test:
Plug in critical points into the second derivative
If POSITIVE there is a min
If NEGATIVE there is a max
(((AGAIN, NOTICE THAT FOR THE SHORTCUT WHETHER THE PLUG IN IS POSITIVE OR NEGATIVE HAS A REVERSED MEANING)))
Using Calculus how can you find a max or a min?
ReplyDelete-There are many different ways to find a max or min in Calculus. All of the one’s we learned so far were in chapter three. To find max or min you have to take the derivative, just like everything else in Calculus. One way is using the first derivative test.
Steps for the first derivative test:
tep 1: Take the derivative.
Step 2: Set it equal to zero.
Step 3: Solve for x.
Step 4: Set up intervals. (-infinity, _ ) u ( _, infinity)
Step 5: Plug a value on the interval into the derivative. If you get a positive number, the function is increasing. If you get a negative number, the function is decreasing.
Step 6: Figure out the max and min. (A max is increasing then decreasing and a min is decreasing then increasing.)
Another way is finding the Critical number, etc. is the same as finding a horizontal tangent. These will also give you a max and min. To find the max and min, you take the derivative, then set equal to zero. If you have an interval and it asks for the max and min, plug your x-values you get from solving the derivative and your endpoints from the interval into the original equation. The biggest number is your max and the smallest number is your min.
Ex. Find the max and min on the graph f(x) = cos π x on the interval of [0, 1/6].
f ’(x) = (-sin π x) (π) = -π sin π x
-π sin π x = 0
sin π x = 0
π x = sin^-1(0)
π x = 0, π, 2π
x = 0, 1, 2
f(0) = cos π (0) = cos 0 = 1
f(1/6) = cos π (1/6) = cos π/6 = ½
f(1) = cos π (1) = cos π = -1
f(2) = cos π (2) = cos 2π = 1
max = 1
min = -1
**HELPFUL WORDS:
Extreme values – max and min
Absolute max and min – the highest and lowest point on an interval
Relative max and min – any max or min not on an interval
Critical number – any max or min (often referred to as x = c)
One way to find a max/min using Calculus is the First Derivative Test:
ReplyDeleteStep 1: Take the derivative.
Step 2: Set it equal to zero.
Step 3: Solve for x.
Step 4: Set up intervals. (-infinity, _ ) u ( _, infinity)
Step 5: Plug a value on the interval into the derivative. If you get a positive number, the function is increasing. If you get a negative number, the function is decreasing.
Step 6: Figure out the max and min. (A max is increasing then decreasing, and a min is decreasing then increasing.)
Ex. Find the max and min of f(x) = x^3 – 3/2x^2.
1) 3x^2 – 3x = 0
2) 3x (x – 1) = 0
3) x = 0, 1
4) (-infinity, 0) u (0, 1) u (1, infinity)
5) f’(-1) = 3(-1)^2 – 3(-1) = 6 increasing
f’(1/2) = 3(1/2)^2 – 3(1/2) = -3/4 decreasing
f’(2) = 3(2)^2 – 3(2) = 6 increasing
6) x = 0 is a max & x = 1 is a min
*Note: If they ask for absolute extrema, you have to plug back into the original equation to find the y-values in order to find the critical points and end points.
**Soooo, how can you find a max or a min?
ReplyDeleteHmm..first let's refresh our memories on section 3.2, shall we? Well in that section we were introduced to Rolle's Theorem. By applying that theorem, it tells you IF there is a max or min on an interval, and it guarantees that you have AT LEAST 1 max or min..Let's not get that confused with the First Derivative Test though. The First Derivative Test is where you actually can find the max's/mins.
-To do that, you first take the derivative of the function.
-Then you set it equal to zero and solve for x (remember, if it happens to be a fraction, you just set the numerator equal to zero and solve)
-Once you get your x-value(s), you check for differentiability.
-Then you set up intervals starting with negative infinity and ending with infinity
-Then you plug values within those intervals into the 1st derivative..to see if you get a positive or negative value (if positive= INCREASING; if negative=DECREASING)
**How to determine if it's a max/min..
-If you have a change going from decreasing to increasing, (neg. to pos.) that means you have a min at whatever x-value it is.
-If you have a change going from increasing to decreasing (pos. to neg.) that means you have a max at whatever the x-value is.
-If you happen to get increasing followed by increasing, then it's nothing..there are no max's or mins. That also goes for if you have decreasing followed by decreasing..
Finding Max or Min using calculus
ReplyDelete1. Take the derivative
2. Set the derivative = 0 and solve for x. Check for differentiability after
3. Set up your intervals from -infinity to infinity
4. Plug in values from your interval. Positive results mean it is increasing and negative results are decreasing
5. If the value changes from decreasing to increasing, it is a min on that point.
If the value changes from increasing to decreasing it is.... you guessed it! a max on that point.
how can you use calculus to find a max or min?
ReplyDeleteuse the first derivative test!
1. take your derivative and set = 0.
2. check your differentiability.
3. set up intervals from negative infinity to infinity
4. plug in the values from the interval. postitive = increase negative = decrease
5. if the value decreases then increases, its a min; if the value increases then decreases its a max.
To find a max or min using calculus, you just use the first derivative test.
ReplyDeletefirst, you take the first derivative.
second, you set the derivative equal to zero
third, you solve for x.
fourth, you set up intervals (-infinity, x)u(x, infinity)
then, you plug numbers from each interval into the first derivative.
if its positive, it is increasing.
if it is negative, then it is decreasing.
if the function is increasing, then decreasing, you have a max.
if it is decreasing then increasing, you have a min.
finally, you state the x-intercept that is your max or min (x=# is a max/min by the first derivative test).
simple as that.
Using Calculus how can you find a max or a min? How do you determine if the value is a max or min?
ReplyDeleteTo find a max or a min of a function using calculus you must use the first derivative test. to use the first derivative test you must first take the derivative of the equation of the function. The you must solve the derivative for x. the answers you get are your critical points. your critical points are the maxs or mins of the function. to find whether a critical point is a max or a min then you must set up the intervals the function using the critical points. plug in a number within the interval into the derivative of the equation. if you get a positive number then that critical point is a max, if you get a negative then that critical point is a min.
There are different ways to find a max or min in Calculus.
ReplyDeleteOne simple way to find maxes or mins is the first derivative test.
1. Find first derivative.
2. Set it equal to zero.
3. Solve for x.
4. Set up intervals for which necessary.
5. Plug in numbers between each number on each interval to the first derivative.
6. If the outcome is positive, it is increasing. If the outcome is negative, it is decreasing.
7. Finally you have to find where it increasing to decreasing, or decreasing to increasing. If it is increasing to decreasing, the point of that is said to be a max. If it is decreasing, the point of that is said to be a min.
Finding a max or min in calculus is pretty simple. All you have to do is use the 1st derivative test. Steps for the first derivative test are:
ReplyDelete1. Take the first derivative
2. Set it equal to zero and solve for x.
3. Check to see if it is differentiable.
4. Set up intervals for the numbers you found from steps 2 and 3.
5. Plug in numbers on each interval.
6. If the solutions from step 5 come out positive then it is increasing, if negative then it is decreasing.
7. Maxs are at points where the interval switches from increasing to decreasing, and mins are at the points where it goes from decreasing to increasing.
Use Calculus and find a max or a min. Easy.
ReplyDeleteThere are more than one ways to find a max or min in Calculus. To find max or min you have to take the derivative, just like everything else in Calculus. Im gonna show you the first derivative test.
Steps for the first derivative test:
1: Take the derivative.
2: Set it equal to zero.
3: Solve for x.
4: Set up intervals. (-infinity, _ ) u ( _, infinity)
5: Plug a value on the interval into the derivative. If you get a positive number, the function is increasing. If you get a negative number, the function is decreasing.
6: Figure out the max and min. (A max is increasing then decreasing and a min is decreasing then increasing.)
Another way is finding the Critical number or something else. These will also give you a max and min. To find the max and min, you take the derivative, then set equal to zero. If you have an interval and it asks for the max and min, plug your x-values you get from solving the derivative and your endpoints from the interval into the original equation. The biggest number is your max and the smallest number is your min.
well, there's a basis for this simple answer: derivatives
ReplyDeleteand you can find out if a # is a max or min by plugging in numbers from an interval and seeing whether the outcome is positive or negative