1. ## Homework questions

Any help is greatly appreciated

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.

x= -3 is a relative min
x = 1.6 is a relative max??

2. Originally Posted by Sammyj
Any help is greatly appreciated

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.

x= -3 is a relative min
x = 1.6 is a relative max??
Show us some of your working did you use the fact that f(x) is concave up or down where the second derivative is positive and negative respectively?

3. I'm not sure what you're asking. If you want me to draw graphs I do not have that capability online.

4. Originally Posted by Sammyj
I'm not sure what you're asking. If you want me to draw graphs I do not have that capability online.
Why not find the first derivative and make this five times easier? Consider that $\displaystyle f'(x)=\int{f''(x)}dx=x^3+2x^2-15x+C$ Now to find C...just use what values make the first derivative zero...in other words the critical poitns

5. 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

Like that?

6. Are questions a & b correct?