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=5cubroot(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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