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) = 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
I still got it wrong even though I found the g''(x) and your answer
Our answer is correct
To find the second derivative, g''(x), you need to differentiate the first derivative, g'(x), with respect to x. Let's start by finding the first derivative, g'(x), of the function g(x):
g(x) = 6x^3 + 72x^2 + 270x
To find g'(x), we will use the power rule for differentiation. This rule states that for a term of the form ax^n, the derivative with respect to x is given by nx^(n-1), multiplied by the coefficient a. Applying this rule to each term in g(x), we get:
g'(x) = 3 * 6x^(3-1) + 2 * 72x^(2-1) + 270 * 1 * x^(1-1)
= 18x^2 + 144x + 270
Now that we have g'(x), we can proceed to find the second derivative, g''(x). Following the same power rule for differentiation, we differentiate g'(x) with respect to x:
g''(x) = 2 * 18x^(2-1) + 144 * 1 * x^(1-1)
= 36x + 144
To evaluate g''(−5), substitute x = -5 into the equation for g''(x):
g''(−5) = 36 * (-5) + 144
= -180 + 144
= -36
Therefore, g''(−5) = -36.