Question

In Exercises \(33-38\) , use the Second Derivative Test to find the local extrema for the function. $$y=x^{5}-80 x+100$$

Short answer

The function \(y = x^{5}-80x+100\) has a local minimum at \(x = 2\) and a local maximum at \(x = -2\).

Step-by-step solution

Finding the First Derivative

The first derivative of the function \(y = x^{5}-80x+100\) is computed using the power rule, which gives \(y' = 5x^{4}-80\).

Finding the Critical Points

Setting the first derivative equal to zero and solving for x will give the critical points. \(0 = 5x^{4}-80\) yields \(x^{4} = 16\) so \(x = \sqrt[4]{16}\). Therefore, the critical points are \(x = 2\) and \(x = -2\).

Finding the Second Derivative

The second derivative of the function is obtained by differentiating the first derivative. This gives \(y'' = 20x^{3}\).

Applying the Second Derivative Test

Substituting the critical points into the second derivative, we get \(y''(2) = 20(2^{3}) = 160\) which is positive, hence by the Second Derivative Test, \(x = 2\) is a local minimum. Similarly, \(y''(-2) = 20(-2^{3}) = -160\) which is negative, hence by the Second Derivative Test, \(x = -2\) is a local maximum.