Suppose that f(x) = (x + 4) (x - 5)2. List all critical numbers of f. If there a

Question

Suppose that f(x) = (x + 4) (x - 5)2. List all critical numbers of f. If there are no critical numbers, enter 'NONE'. Critical numbers = Use interval notation to indicate where f (x) is increasing. Note: Use 'INF' for infinity, '-INF' for -infinity, and use 'U' for the union symbol. Increasing: Use interval notation to indicate where f (x) is decreasing. Decreasing: List the x-coordinates of all local maxima of f. If there are no local maxima, enter 'NONE'. X values of local maxima = List the x-coordinates of all local minima of f. Note: If there are no local minima, enter 'NONE'. X values of local minima = Use interval notation to indicate where f (x) is concave up. Concave up: Use interval notation to indicate where f (x) is concave down. Concave down: List the x values of all inflection points of f. If there are no inflection points, enter 'NONE'. X values of inflection points = Use all of the preceding information to sketch a graph of f. When you're finished, enter a "1" in the box below. You may need to turn in your sketch in class. Graph Complete:

Explanation / Answer

f(x) =(x+4)(x-5)^2

f'(x) = 0

f'(x) = (x-5)^2 + (x+4)*2*(x-5) = 0

=> (x-5) (x-5+2x+8) = 0

=>x= 5 and x= -1

critical points are x= -1,5

f"(x) = 2(x-5) + 2(x-5) +2(x+4) = 4(x-5) +2(x+4)

at x=-1 , f"(x) = -12-12+6 < 0

f(x) has local max at x=-1

at x=5 ,f"(x) = 0+0+18 > 0

f(x) has local min at x=5

maximum value , f(-1) = (-1+4) (-1-5)^2 = 3* 36 = 108

minimum value ,f(5) = (5+4) (5-5) = 0

To find INFLECTION points , f"(x) = 0

4(x-5)+2(x+4) =0

=> 6x - 12 =0

inflection point , x= 2

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