To determine concavity, you use the same steps as the first derivative test, but with the 2nd derivative. To review, here are the steps: 1. Take the 2nd derivative. 2. Set it equal to 0. 3. Solve for x. 4. Check differentiability. 5. Set up intervals. 6. Pick numbers on the intervals and plug them into the 2nd derivative. 7. If you get a positive number, it is concave up. If you get a negative number it is concave down. 8. If there is a change (like from positive to negative or vice versa), it is a point of inflection.
Ex. 1) Determine the open intervals on which the graph f(x) = -x^3 + 6x^2 – 9x – 1 is concave up or down. f’(x) = -3x^2 + 12x – 9 f’’(x) = -6x + 12 -6x + 12 = 0 -6x = -12 x = 2 It is differentiable. (-infinity, 2) u (2, infinity) f’’(1) = positive # f’’(3) = negative # concave up: (-infinity, 2) concave down: (2, infinity) *Note that the question does not ask for the point of inflection, but if it did, you would say: At x = 2 there is a point of inflection.
**Determining cocavity--Is it concave up or down? -First off, when you see words like "discuss the concavity of..", "find all points of inflection of..", or "determine the open intervals on which the graph is concave upward or concave downward", the first thing that should pop into your mind is SECOND DERIVATIVE. -The 2nd derivative test is very similar to the 1st derivative test..the only difference being that you have to take the 1st derivative of the function, then take the 2nd derivative, and then set the 2nd derivative equal to zero and solve for x..Also, in the 2nd derivative test you're not determining whether points on the graph are increasing or decreasing. You're finding where the graph is concave up or concave down..all that means is after you would set up your intervals, you would plug numbers from those intervals into the 2nd derivative to get either a positive or negative number..if positive that means it's concave up; if negative, that means it's concave down. -Points of inflection are x-values where the change goes from negative to positve, or positive to negative (once you plug in your numbers)..So if you get negative followed by negative for an x-value, that means that x-value is not a point of inflection
Here's an example:
1.) Determine the points of inflection and discuss the concavity of the graph of the function: f(x)=x^3-9x^2 *So first you want to take the 1st derivative of this function and you should get 3x^2-18x...butttt don't stop there! Remember* "concavity" infers "2nd derivative test" *So next you're going to take the 2nd derivative of the function and you should end up with: 6x-18 *Now you can set that equal to zero and solve for x like this: 6x-18=0 6x=18 x=3 *Now set up your intervals with 3 (-infinity,3)u(3,infinity) *Now plug in numbers within those intervals into the 2nd derivative (not the 1st) *For the first interval you can plug in 0 and you should get a negative number. For the second interval you can plug in 4 and you should get a positive number *Soooooo, since you have a change that's going from positive to negative that means x=3 is a point of inflection and the graph of the function is concave down on (-infinity,3) and it's concave up on (3,infinity).
To determine Concavity you must perform the second derivative test. STEPS TO USING THE SECOND DERIVATIVE TEST:
Step #1: take the first derivative set equal to zero and solve. ((keep in mind that this step will not always be necessary but can always be performed anyway. This step is not necessary when one looking for concavity or points of inflection.))
Step #2: take the second derivative set = 0 and solve
Step #3: Check differentiability
Step #4: set up intervals
Step #5: pick numbers on the interval and plug them into the SECOND derivative
Step #6: if POSITIVE: the function is concave up on that interval if NEGATIVE: the function is concave down on that interval
Example #1:
Determine the open intervals on which the graph is concave up or concave down: F(x)= 6/ x^2+3
Step #1: take the first derivative set equal to zero and solve. (( But this step is not necessary because the problem is looking for concavity.))
Step #2:
take the second derivative set = 0 and solve
(x^2+3)(0)-[(6)(2x)]/ (x^2+3)^2
F'(x)= -12x/ (x^2+3)^2
(x^2+3)^2(-12)-[(-12x)(2(x^2+3)(2x))]/ (x^2+3)^2
(x^2+3)(-12)+48x^2
-12x^2-36+48x^2
F"(x)= -36x^2-36=0
-36x^2=36
x^2= 1
x= +/- 1
Step #3: Check differentiability
DIFFERENTIABLE? yes.
Step #4: set up intervals
(-infinity,-1)U(-1,1)U(1, infinity)
Step #5: pick numbers on the interval and plug them into the SECOND derivative
F"(-2)= 36(-2)^2-36/ (-2^2+3)^2 = POSITIVE
F"(0)= 36(0)^2-36/ (0^2+3)^2 = NEGATIVE
F"(2)= 36(2)^2-36/ (2^2+3)^2= POSITIVE
Step #6: if POSITIVE: the function is concave up on that interval
Therefore the function is concave up on the intervals (-infinity, -1)U(1,Infinity)
if NEGATIVE: the function is concave down on that interval
Therefore the function is concave down on the interval (-1,1)
TO see how to tell if the interval is concave up or concave down we can look at step six of the second derivative test steps
this step reads "if POSITIVE: the function is concave up on that interval if NEGATIVE: the function is concave down on that interval"
Meaning that whn you plug the points on the intervals into the second derivative wheter the outcome is positive or negative will determine whether the interval is concave up or concave down.
if the outcome is positive the interval is concave up if the outcome is negative the interval is concave down
EXAMPLE of how to determine whether the interval is concave up or concave down
pick numbers on the interval and plug them into the SECOND derivative
F"(-2)= 36(-2)^2-36/ (-2^2+3)^2 = POSITIVE
F"(0)= 36(0)^2-36/ (0^2+3)^2 = NEGATIVE
F"(2)= 36(2)^2-36/ (2^2+3)^2= POSITIVE
Step #6: if POSITIVE: the function is concave up on that interval
Therefore the function is concave up on the intervals (-infinity, -1)U(1,Infinity)
if NEGATIVE: the function is concave down on that interval
Therefore the function is concave down on the interval (-1,1)
How do you know if it is concave up or down? The process used to find out where a graph is concave up and concave down is almost identical to the process that is used to find critical numbers. The only difference is you are working with the second derivative.
How do you determine concavity? The first step in determining concavity is to take the first derivative. Next you’re going to take the second derivative. Once that’s done your going to set the second derivative equal to zero and solve for x. Using the values you get for x, set up an interval to test the concavity. To find whether the graph is concave up or down, plug in numbers on that interval into the first derivative. If the number is positive it is a max, if it is negative, it’s a min.
1. take the 1st derivative test 2. find increasing/decreasing 3. take the second derivative test 4. set = 0 5. plug into the 1st derivative 6. positve numbers are maxes, negative numbers are mins. positive is concave up, negative is down
how do you determine concavity? How do you know if it is concave up or down?
you must first take the derivative of the equation. then you must take the second derivative. you must then find your critical numbers. use must set the derivative equal to zero. then you must set up your intervals. take a number out of each interval and plug it into the second derivative if you get a negative number, that interval is concave up. if it is a positive number then it is concave down.
take the first derivative, set up intervals, then the second derivative... then set the function = 0. plug in your answer back into the first derivative; positive numbers are maxes and negatives are mins. positives = concave up negatives = concave down
To determine concavity, you use the same steps as the first derivative test, but with the 2nd derivative. To review, here are the steps:
ReplyDelete1. Take the 2nd derivative.
2. Set it equal to 0.
3. Solve for x.
4. Check differentiability.
5. Set up intervals.
6. Pick numbers on the intervals and plug them into the 2nd derivative.
7. If you get a positive number, it is concave up. If you get a negative number it is concave down.
8. If there is a change (like from positive to negative or vice versa), it is a point of inflection.
Ex. 1) Determine the open intervals on which the graph f(x) = -x^3 + 6x^2 – 9x – 1 is concave up or down.
f’(x) = -3x^2 + 12x – 9
f’’(x) = -6x + 12
-6x + 12 = 0
-6x = -12
x = 2
It is differentiable.
(-infinity, 2) u (2, infinity)
f’’(1) = positive #
f’’(3) = negative #
concave up: (-infinity, 2)
concave down: (2, infinity)
*Note that the question does not ask for the point of inflection, but if it did, you would say: At x = 2 there is a point of inflection.
**Determining cocavity--Is it concave up or down?
ReplyDelete-First off, when you see words like "discuss the concavity of..", "find all points of inflection of..", or "determine the open intervals on which the graph is concave upward or concave downward", the first thing that should pop into your mind is SECOND DERIVATIVE.
-The 2nd derivative test is very similar to the 1st derivative test..the only difference being that you have to take the 1st derivative of the function, then take the 2nd derivative, and then set the 2nd derivative equal to zero and solve for x..Also, in the 2nd derivative test you're not determining whether points on the graph are increasing or decreasing. You're finding where the graph is concave up or concave down..all that means is after you would set up your intervals, you would plug numbers from those intervals into the 2nd derivative to get either a positive or negative number..if positive that means it's concave up; if negative, that means it's concave down.
-Points of inflection are x-values where the change goes from negative to positve, or positive to negative (once you plug in your numbers)..So if you get negative followed by negative for an x-value, that means that x-value is not a point of inflection
Here's an example:
1.) Determine the points of inflection and discuss the concavity of the graph of the function:
f(x)=x^3-9x^2
*So first you want to take the 1st derivative of this function and you should get 3x^2-18x...butttt don't stop there! Remember* "concavity" infers "2nd derivative test"
*So next you're going to take the 2nd derivative of the function and you should end up with:
6x-18
*Now you can set that equal to zero and solve for x like this:
6x-18=0
6x=18
x=3
*Now set up your intervals with 3
(-infinity,3)u(3,infinity)
*Now plug in numbers within those intervals into the 2nd derivative (not the 1st)
*For the first interval you can plug in 0 and you should get a negative number. For the second interval you can plug in 4 and you should get a positive number
*Soooooo, since you have a change that's going from positive to negative that means x=3 is a point of inflection and the graph of the function is concave down on (-infinity,3) and it's concave up on (3,infinity).
To determine Concavity you must perform the second derivative test.
ReplyDeleteSTEPS TO USING THE SECOND DERIVATIVE TEST:
Step #1:
take the first derivative set equal to zero and solve. ((keep in mind that this step will not always be necessary but can always be performed anyway. This step is not necessary when one looking for concavity or points of inflection.))
Step #2:
take the second derivative set = 0 and solve
Step #3:
Check differentiability
Step #4:
set up intervals
Step #5:
pick numbers on the interval and plug them into the SECOND derivative
Step #6:
if POSITIVE: the function is concave up on that interval
if NEGATIVE: the function is concave down on that interval
Example #1:
Determine the open intervals on which the graph is concave up or concave down:
F(x)= 6/ x^2+3
Step #1:
take the first derivative set equal to zero and solve. (( But this step is not necessary because the problem is looking for concavity.))
Step #2:
take the second derivative set = 0 and solve
(x^2+3)(0)-[(6)(2x)]/ (x^2+3)^2
F'(x)= -12x/ (x^2+3)^2
(x^2+3)^2(-12)-[(-12x)(2(x^2+3)(2x))]/ (x^2+3)^2
(x^2+3)(-12)+48x^2
-12x^2-36+48x^2
F"(x)= -36x^2-36=0
-36x^2=36
x^2= 1
x= +/- 1
Step #3:
Check differentiability
DIFFERENTIABLE?
yes.
Step #4:
set up intervals
(-infinity,-1)U(-1,1)U(1, infinity)
Step #5:
pick numbers on the interval and plug them into the SECOND derivative
F"(-2)= 36(-2)^2-36/ (-2^2+3)^2 = POSITIVE
F"(0)= 36(0)^2-36/ (0^2+3)^2 = NEGATIVE
F"(2)= 36(2)^2-36/ (2^2+3)^2= POSITIVE
Step #6:
if POSITIVE: the function is concave up on that interval
Therefore the function is concave up on the intervals (-infinity, -1)U(1,Infinity)
if NEGATIVE: the function is concave down on that interval
Therefore the function is concave down on the interval (-1,1)
TO see how to tell if the interval is concave up or concave down we can look at step six of the second derivative test steps
this step reads
"if POSITIVE: the function is concave up on that interval
if NEGATIVE: the function is concave down on that interval"
Meaning that whn you plug the points on the intervals into the second derivative
wheter the outcome is positive or negative will determine whether the interval is concave up or concave down.
if the outcome is positive the interval is concave up
if the outcome is negative the interval is concave down
EXAMPLE of how to determine whether the interval is concave up or concave down
pick numbers on the interval and plug them into the SECOND derivative
F"(-2)= 36(-2)^2-36/ (-2^2+3)^2 = POSITIVE
F"(0)= 36(0)^2-36/ (0^2+3)^2 = NEGATIVE
F"(2)= 36(2)^2-36/ (2^2+3)^2= POSITIVE
Step #6:
if POSITIVE: the function is concave up on that interval
Therefore the function is concave up on the intervals (-infinity, -1)U(1,Infinity)
if NEGATIVE: the function is concave down on that interval
Therefore the function is concave down on the interval (-1,1)
How do you know if it is concave up or down?
ReplyDeleteThe process used to find out where a graph is concave up and concave down is almost identical to the process that is used to find critical numbers. The only difference is you are working with the second derivative.
How do you determine concavity?
The first step in determining concavity is to take the first derivative. Next you’re going to take the second derivative. Once that’s done your going to set the second derivative equal to zero and solve for x. Using the values you get for x, set up an interval to test the concavity. To find whether the graph is concave up or down, plug in numbers on that interval into the first derivative. If the number is positive it is a max, if it is negative, it’s a min.
Example:
Y=5x^3+9x^2-5x+6
Y’=15x^2+18x-5
Y”=30x+18
30x=18
X=-3/5
Concave down:(-infinity,-3/5) Concave up:(-3/5,infinity)
1. take the 1st derivative test
ReplyDelete2. find increasing/decreasing
3. take the second derivative test
4. set = 0
5. plug into the 1st derivative
6. positve numbers are maxes, negative numbers are mins. positive is concave up, negative is down
how do you determine concavity? How do you know if it is concave up or down?
ReplyDeleteyou must first take the derivative of the equation. then you must take the second derivative. you must then find your critical numbers. use must set the derivative equal to zero. then you must set up your intervals. take a number out of each interval and plug it into the second derivative if you get a negative number, that interval is concave up. if it is a positive number then it is concave down.
how to determine concavity:
ReplyDeletetake the first derivative, set up intervals, then the second derivative... then set the function = 0.
plug in your answer back into the first derivative; positive numbers are maxes and negatives are mins. positives = concave up negatives = concave down