2nd Derivative is f"(x)= 3x^2 + 4x -15, critical pts are x =

sammyj

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Any help is greatly appreciated. I'm not sure if I'm doing this correctly :?

Use the 2nd derivative test when answering these questions. The critical points of the function f(x) are x = -5, x = 0 and x = 3. The first derivative f'(x) is not known. But the second derivative is f"(x)= 3x^2 + 4x -15

a) Find all the intervals where the function f(x) is increasing and all the intervals where the function f(x) is decreasing. Explain your reasoning.

x =-5 3(-5)^2 + 4(-5) -15 = -110 Decreasing (-oo, -5)?
x = 0 3(0)^2 + 4(0) - 15 = -15 Decreaing (-5, 0)?
x = 3 3(3)^2 + 4(3) - 15 = 24 Increasing (0, 3)?

b) Find all the intervals where the function f(x) is concave up and all the intervals where the function f(x) is concave down. Explain your reasoning.

x =-4 3(-4)^2 + 4(-4) -15 = -79 Concave Down (-oo, -3)
x = 0 3(0)^2 + 4(0) - 15 = -15 Concave Down (-3, 1.6)
x = 2 3(2)^2 + 4(2) - 15 = 5 Concave Up (1.6, oo)

c)Identify the points where all the relative minima occur and all the points where the relative maxima occur. Explain your reasoning.

Critical points/ F"(x) = 3x^2 + 4x - 15 / F(x) is concave
______________________________________________

-5 / 30-20-15= -5 NEG / Down x = -5 is a max
0 / -15 NEG / Down x = 0 is a max
3 / 18 + 12 - 15 = 15 POS / Up x = 3 is max
 
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