Posted by **Natalie** on Tuesday, July 22, 2014 at 6:36pm.

For f(x)=2(x+5)^3 +7

Find and classify the extreme values, determine where the function is increasing and decreasing, where it is concave up and concave down, and any points of inflection.

I understand this is using the first and second derivative tests, but I can't find any zeroes for the derivative of this function. Am I doing something wrong?

- Calculus -
**bobpursley**, Tuesday, July 22, 2014 at 6:45pm
f(x)=0=2(x+5)^3 +7

(x+5)^3=-7/2

x+5= -cubroot(7/2)

x=-5-cubroot(7/2)

Those are the zeroes, but I don't see in the problem wording it asked for a zero

- Calculus -
**Damon**, Tuesday, July 22, 2014 at 6:47pm
y = 2 (x+5)^3 + 7

to make this really easy you can say

z = x+5

then dz/dx = 1

y = 2 z^3 + 7

dy/dz = 6 z^2

d^2y/dz^2 = 12 z

so

dy/dx = dy/dz * dz/dx = 6(x^2+10x+25)

when is that zero ?

(x+5)(x+5) = 0

x = -5 twice

what is d/dx (dy/dx)?

6 ( 2 x + 10)

when x = -5

2(-5)+10 = 0

so that is not a max or a min but an inflection point.

- Calculus -
**Steve**, Tuesday, July 22, 2014 at 7:27pm
Think of the graph for y = x^3

It has no max/min, but just an inflection point at x=0.

You function is just the same, only stretched by 2, shifted left by 5, and shifted up by 7.

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