Question

Given the function g(x)=4x 3 +24x 2 +36x, find the first derivative, g ′ (x). g ′ (x)= Notice that g (x)=0 when x=−3. That is, g ′ (−3)=0. Now, we want to know whether there is a local minimum or local maximum at x=−3, so we will use second derivative test. Find the second derivative, g ′′ (x). g "(x)= Evaluate g′′ (−3) g ′′(−3)= Based on the sign of this number, does this mean the graph of g(x) is concave up or concave down at x=−3 ? At x=−3 the graph of g(x) is concave down concave up Based on the concavity of g(x) at x−3, does this mean that there is a local minimum or local maximum at x=−3 ? At x=−3 there is a local minimum maximum

Answer 1

The function given is g(x) = 4x^3 + 24x^2 + 36x.

The first step to solve this problem is to find the first derivative of g(x), denoted as g'(x).
In this case, the first derivative becomes g'(x) = 12x^2 + 48x + 36.

Determining where g'(x) equals 0 can help identify any potential local minimums or maximums. It turns out that g'(x) equals 0 at x equals -3.

To establish whether this point is a local minimum or a maximum, we use the second derivative test. We need to calculate the second derivative of g(x), denoted as g''(x).

The result in our case is g''(x) = 24x + 48.

Then, you can evaluate the second derivative at x equals -3. When substituting "-3" into the second derivative, g''(-3) equals -24.

Considering the result of the second derivative, it is concluded that the graph of g(x) at x=-3 is concave down because g''(-3) is less than 0. According to this concave down nature, it suggests that there is a local maximum at x = -3.

So, summarizing all these findings - the first derivative, g'(x) is 12x^2 + 48x + 36, the second derivative, g''(x) is 24x + 48, the second derivative evaluated at -3, g''(-3) is -24, and based on it, it's confirmed that the graph of g(x) is concave down and therefore has a local maximum at x = -3.