Question

In Exercises \(33-38\) , use the Second Derivative Test to find the local extrema for the function. $$y=3 x-x^{3}+5$$

Short answer

The local minimum is at \(x = -1\) and the local maximum is at \(x = 1\).

Step-by-step solution

Find the First Derivative

Find the first derivative of the function \(y\). The derivative of a function gives the rate of change of the function at any point. For the function \(y = 3x - x^3 + 5\), the first derivative, \(y'\), can be found using basic rules of differentiation. The derivative of \(3x\) is 3, the derivative of \(-x^3\) is \(-3x^2\), and the derivative of 5 is 0. Therefore, \(y' = 3 - 3x^2\).

Find the Critical Points

The critical points occur where the first derivative is zero or undefined. This function's derivative is defined for all \(x\), so set the first derivative equal to zero and solve for \(x\).\n3 - 3x^2 = 0\nSolving for \(x\) gives \(x = -1, 1\). These are the critical points.

Find the Second Derivative

Make the derivative of \(y'\) to find the second derivative \(y''\). The second derivative is essential in determining whether a critical point is a maximum, minimum, or an inflection point. The derivative of \(3 - 3x^2\) is \(-6x\).

Classify the Critical Points

Apply the Second Derivative Test. Plug the critical points into the second derivative. If the result is positive, the point is a local minimum, if negative, it’s a local maximum, if zero, the test is inconclusive. Plugging in \(x = -1\) into \(-6x\) gives 6, a positive number, so \(x = -1\) is a local minimum. Plugging in \(x = 1\) gives -6, a negative number, so \(x = 1\) is a local maximum.