Kuan Liu, Ryan Park, Nathan Saedi, Sabrina Sauri & Ellie Tsang Function Analysis Kuan Liu, Ryan Park, Nathan Saedi, Sabrina Sauri & Ellie Tsang

Function Analysis Used when you have a function f(x) and you want to find: Intervals where f(x) is increasing Intervals where f(x) is decreasing Intervals where f(x) is concave up Intervals where f(x) is concave down Values of x where there is a relative minimum Values of x where there is a relative maximum Values of x where there is a point of inflection

Mean Value Theorem MVT states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that. Remember that in order for The Mean Value Theorem for Derivatives to work, the curve must be continuous on the interval and at the endpoints.

Example To apply the Mean Value Theorem to the function f(x)= x^2 -5x+7 , -1<x<3 We first calculate the quotient f(3)-f(-1) 1-13 ------------ = ----------- = -3 3-(-1) 4 Next we take the derivative f’(x)=2x-5 and equate it to the result of the calculation above: 2x-5=-3 Then solve this equation to get x=1

Increasing and Decreasing Functions How to determine on what intervals a function f(x) is increasing and/or decreasing 1. Take the first derivative of the function. 2. Set the first derivative equal to zero. 3. Solve for x to find critical points Critical point: a point on a function f(x) in which the derivative is 0 4. Complete a line test. Plug in numbers smaller and larger than each critical point into the first derivative. If f’(x) > 0 for all x in (a, b), then f is increasing on [a, b]. If f’(x) < 0 for all x in (a, b), then f is decreasing on [a, b]. If f’(x) = 0 for all x in (a, b), then f is constant on [a, b].

Example Find the open intervals on which f(x) = 2x3 - 3x2 is increasing or decreasing. f’(x) = 6x2 - 6x 6x2 - 6x = 0 6x(x - 1) = 0 → x = 0, 1 + - + 1 f’(x) = 6x2 - 6x f’(-1) = 6(-1)2 - 6(-1) f’(-1) = 12 f’(-1) > 1 f’(x) = 6x2 - 6x f’(½) = 6(½)2 - 6(½) f’(½) = -3/2 f’(½) < 0 f’(x) = 6x2 - 6x f’(2) = 6(2)2 - 6(2) f’(2) = 12 f’(2) > 0 f(x) is increasing on the intervals ( ∞ , 0) and (1, ∞ ) and decreasing on the interval (0, 1), as shown by the line test.

Relative Maximums and Minimums Maximum and minimum functions are the largest and the smallest value that a function takes at a point The relative maximum and minimum is determined when the slope of the point is = 0 This is found by using the first derivative After you find the first derivative, find the zeroes of the equation to get the Relative maximums or minimums To determine which it is (max or min) plug in a value between, before, and after the zeroes (+, zero, - = max) (-, zero, + =min)

Example Let f(x)= x2 − 6x + 5 Are there any critical values or any turning points? If so, do they determine a maximum or a minimum? And what are the coordinates on the graph of that maximum or minimum? f’(x) = 2x - 6 2x - 6 = 0 Implying x = 3 (Critical Point) f(3) = 32 − 6(3) + 5 f(3) = -4 (Minimum value)

Concavity Concavity is the direction the parabolic function opens up To find concavity you need to find the second derivative of the given function After you find the second derivative, you find the zeroes of the second derived equation to find the inflection point Then plug in a value between the inflection points Example:

Points of Inflection How to find the points of inflection on a function f(x) 1. Take the second derivative of the function 2. Set the second derivative equal to zero. 3. Find x to find the points of inflection Points of Inflection: points where concavity changes on the function 4. Complete a line test Plug in numbers smaller and larger than each Inflection point into the second derivative. If f”(x) > 0 for all x in (a, b), then f is concave up on [a, b]. If f” (x) < 0 for all x in (a, b), then f is concave down on [a, b].

Example f(x)= Find the point of inflection of the function on f(x)=x^(3) -2x+4 f’(x)= 3x^(2) -2 f”(x)= 6x 6x=0 x=0 f” 6(-2)= -12 neg.=concave down 6(2)= 12 pos.=concave up