Given the function g (x) = 6x^3 + 72x^2 + 270x , find the first derivative, g'( x) .

Question

g'( x ) =

Notice that g ' ( x ) = 0 when x = − 5 , that is, g ' ( − 5 ) = 0 . Now, we want to know whether there is a local minimum or local maximum at x = − 5 , so we will use the second derivative test.

Find the second derivative, g ' ' ( x ) .

Evaluate g''(−5).
g''(−5)=

im confused on how to find (Evaluate g''(−5).
g''(−5)=) thats it

Answer 1

g (x) = 6x^3 + 72x^2 + 270x
g' (x) = 18x^2 + 144x + 270
g '' (x) = 36x + 144

They told you that g ' (-5) = 0, I trust them to be correct

g '' (-5) = 36(-5) + 144 = -36
the fact that it is negative is the important part.
It means that when x = -5 you have a maximum of the function

In general, if for some value of x = a, f ' (a) = 0
proceed to find f ''(a).
If f ''(a) > 0 , then you will have a minimum when x = a
If f ''(a) < 0 , then you will have a maximum when x = a
If f ''(a) = 0 , then you will have a point of inflection when x = a

Answer 2

I still got it wrong even though I found the g''(x) and your answer

Answer 3

Our answer is correct