because we will soon be moving on to ap review and i am scared to death im going to expand upon the basics
In that i mean I am going to expand upon that work sheet we were given to help remember what processes we use when solving various types of problems
1. Slope of a normal line: Take a derivative and plug in the x-‐value. Take the negative reciprocal of the number. (not covered)
2. Average rate of change/ average velocity: slope where a,b are x-values
Find the slope of the graph of f(x) = x^2 + 3x + 1 at (1, 3)
lim as Dx approaches 0: ( f(x + Dx) - f(x) ) / (Dx)
What you have to do is plug in (x+ Dx) for every x in the formula.
(x + Dx)^2 + 3(x + Dx) + 1 - (x^2 + 3x + 1) / (Dx)
That simplifies to...
x^2 + 2xDx + Dx^2 + 3x + 3Dx + 1 - x^2 - 3x -1 / Dx
2xDx + Dx^2 + 3Dx / Dx
2x + Dx + 3
plug in 0 for Dx
2x + 3 is your equation
Now just plug in the x coordinate to find the slope.
2(1) + 3 = 5
The slope is 5.
3. rate of change: derivative. if given an x value plug in to get a number
Example:
A Pebble is dropped into a calm pond causing ripples in the form of concentric circles. The radius of the outer circle is increasing at a rate of 1 foot PER second when the radius is 4 ft at what rate is the total area (A) of the disturbed water changing?
#1 dr/dt= 1ft/sec
R=4 ft
dA/dt= ?
#2 A= pi r ^2
#3 dA/dt= pi{2r dr/dt}
#4 dA/dt= 2pi (4)(1)
Therefore 8 pi ft/sec
4. Slope of a horizontal tangent line
take derivative set equal to zero and solve for x
Ex:
Find the slope of the graph of f(x)= 2x-3 at (2,1)
First you would plug in:
f(x) 2(x+&x)-3-(2x-3)/ &x
&x-> 0
Expand:
2x+2&x – 3 – 2x + 3 / &x
Take out what cancels:
2x + 2&x -3 -2x +3/&x
And you’re left with:
2&x/&x
Simplify:
2 &x/ &x
Therefore the answer is 2
5. equation of a tangent line
take the dericative and plug in the x value
if you are not given a y value plug the original equation to get the y=value
then plug these values into point slope formula
EXAMPLE:
Find the equation of the tangent line to the graph of the function f(x)= sqrt x-3 at the point (39,6)
Sqrt x-3
Simplifies to (x-3)^1/2
Take the derivative
d/dx= 1/2( x-3)^-1/2 (1)
convert back to a fraction 1/ 2(x-3)^1/2
plug in for x 1/ 2 (39-3)^1/2
Solve = 1/ 2 (36)^1/2
Simplify= 1/ 2 sqrt 36
Solve= 1/12
Therefore the derivative of f(x)= sqrt x-3 at the point (39,6) is 1/12
Because this is the equation of a tangent line you must plug into the equation of a line formula to answer the question of the problem.
Therefore you will plug into y-y1 = m(x-x1)
plug in= y-6 = 1/12 (x-39)
solve: y-6 = 1/12(x-39)
+6 +6
Y= 1/12x + 11/4
Therefore the equation of the tangent line to the graph of the function f(x)= sqrt x-3 at the point (39,6) is y=1/12x + 11/4
6. average value ((not yet covered..)
7. maximum minimum critical values increasing decreasing
take derivative set equal to zero and solve for x
set up intervals using these numbers
plug in numbers on each interval
if it changes from positive to negative it is a maximum. If it changes from negative to positive it is a minimum
all potential maximums and minimums are called critical values. If the interval is positive the interval is increasing
id the interval is negative then the interval is decreasing
Find any critical numbers of the function f(x) = 4x / x^2+1
*Since they're asking you to find critical numbers, the first thing you have to do is take the derivative of the function
*So, using the quotient rule, you get:
*(x^2+1)(4) - [(4x)(2x)] / (x^2+1)^2
Simplifying that you get:
-4x^2+4 / (x^2+1)^2
*To find the critical numbers, you take the numerator of the fraction and set it equal to zero.
So you get:
-4x^2 + 4 = 0
x^2 = 1
*So your critical numbers are x = 1, -1
8. point of inflection concave up or down
take second derivative set equal to zero and solve for x
set up intervals using these numbers plug in numbers on each interval if the intervalsa change signs then there is a point of inflection at that value if it does not change sign then there is not a point of inflection
if the interval is positive then the interval is concave up is the interval is negative then the interval is concave down
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: x = 2 is a point of inflection.
9. linearization ((have not covered))
10. infinite limits
id the degree of the top is greater than the degree of the bottom the limit is infinity. If the degree of the top is less than the degree of the bottom the limit is zero is the degree of the top is equal to the degree of the bottom the limit is the coefficients of the terms with the highest exponents
ex:
2x/3x therefore 2/3
11.vertical asymptotes
factor the top and the bottom of the fraction make cancellations if possible set what is left in the bottom equal to zero and solve for x
vertical asymptotes are given in the form of x= a number
x^2+2x-3/x^2 -5x -6
x2 – 5x – 6 = 0
(x – 6)(x + 1) = 0
x = 6 or –1
12. removables
factor the top and the bottom of the fraction make cancellations set each thing that cancelled equal to zero and solve for x removables are given in the form of x= a number
13. horizontal asymptotes
take the limit as x goes to infinty of the equation
if you get a number then your answer is y= a number
if you get infinity there are no horizontal asymptotes
14. area if only one equation is given (not covered)
15. area between two equations (not covered)
16. volume by disks (not covered)
17. volume by washers (not covered)
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