Question

In Exercises \(33-38\) , use the Second Derivative Test to find the local extrema for the function. $$y=x^{3}+3 x^{2}-2$$

Short answer

Thus, the function \(y=x^{3}+3 x^{2}-2\) has a local minimum at x=0 and a local maximum at x=-2.

Step-by-step solution

Compute the first derivative

The first derivative of the function is computed as follows: \[y' = 3x^{2}+6x\]. This gives the slope of the tangent line to the function at any point.

Find the critical points

The critical points are where the first derivative equals zero or is undefined. So we need to solve the equation \(3x^{2}+6x=0\) for x. Factoring out a 3x, we get \(3x(x+2)=0\). Setting each factor equal to zero gives x=0, x=-2. So the critical points are at x=0, x=-2.

Compute the second derivative

The second derivative of the function is computed as follows: \[y'' = 6x+6\]. The second derivative provides information about the concavity of the function.

Apply the Second Derivative Test

The Second Derivative Test states that if the second derivative at a point is positive, then the function has a local minimum at that point. If the second derivative at a point is negative, the function has a local maximum at that point. If the second derivative equals zero or is undefined at the point, then the test is inconclusive. Now let's calculate the second derivative at our critical points: \[y''(0) = 6\cdot0+6 = 6 > 0\] This means that x=0 is a local minimum. \[y''(-2) = 6\cdot(-2)+6 = -6 < 0\] This means that x=-2 is a local maximum.