Asked by Joshua
Given the function g (x) = 6x^3 + 72x^2 + 270x , find the first derivative, g'( x) .
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
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
Answers
Answered by
mathhelper
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
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
Answered by
Joshua
I still got it wrong even though I found the g''(x) and your answer
Answered by
mathhelper
Our answer is correct
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