Question

In Exercises \(33-38\) , use the Second Derivative Test to find the local extrema for the function. $$y=3 x^{5}-25 x^{3}+60 x+20$$

Short answer

The local extrema of the function \(y=3x^{5}-25x^{3}+60x+20\) are at \(x=-2\) (local maximum), \(x=-1\) (local minimum), \(x=1\) (local maximum), and \(x=2\) (local minimum).

Step-by-step solution

Compute the first derivative

The first derivative of the function is given by differentiating it with respect to x. That will be \(y'=15x^{4}-75x^{2}+60\).

Find the critical points

The critical points are found by setting the first derivative \(y'\) equal to zero and solving for \(x\). That will give \(15x^{4}-75x^{2}+60 = 0\). Factoring out 15 gives \(x^{4}-5x^{2}+4 = 0\). By letting \(z=x^{2}\), the equation turns to \(z^{2}-5z+4 = 0\). This factors to \((z-4)(z-1) =0 \). So, \(z=1\) or \(z=4\). Re-substituting \(z=x^{2}\) gives possible critical points as \(x = -2, -1, 1, 2\).

Compute the second derivative

The second derivative of the function is given by differentiating \(y'\) with respect to x. That will be \(y''=60x^{3}-150x\).

Apply the Second Derivative Test

Substitute the critical points into \(y''\). For \(x=-2\), \(y''=(-2400)<0\), hence \(x=-2\) is a local maximum. For \(x=-1\), \(y''=(210)>0\), hence \(x=-1\) is a local minimum. For \(x=1\), \(y''=(-210)<0\), hence \(x=1\) is a local maximum. For \(x=2\), \(y''=(2400)>0\), hence \(x=2\) is a local minimum. Hence, the local extrema are at \(x=-2, -1, 1, 2\).