Question

Determine the open intervals on which the graph is concave upward or concave downward. \(y=\frac{-3 x^{5}+40 x^{3}+135 x}{270}\)

Short answer

The graph of \(y=\frac{-3 x^{5}+40 x^{3}+135 x}{270}\) is concave upward on the intervals \(-\sqrt{5}\sqrt{5}\), and concave downward on the intervals \(x<-\sqrt{5}\) and \(0<x<\sqrt{5}\).

Step-by-step solution

Find the first derivative

The first derivative of \(y\) is given by \(y'=\frac{d}{dx} \left(\frac{-3 x^{5}+40 x^{3}+135 x}{270}\right)\). This simplifies to \(y'=\frac{-15x^{4}+120x^{2}+135}{270}\).

Find the second derivative

Next, differentiate \(y'\) to find the second derivative. \(y''=\frac{d}{dx} \left(\frac{-15x^{4}+120x^{2}+135}{270}\right)\) simplifies to \(y''=\frac{-60x^{3}+240x}{270}\).

Solve for critical points

To find the critical points of the graph, set \(y''=0\) and solve for \(x\). The solutions, \(x=0\) and \(x=\pm\sqrt{5}\), represent points where the graph might change concavity.

Determine concavity

The graph is concave upward where \(y''>0\), and concave downward where \(y''<0\). Using the critical points, \(x<-\sqrt{5}\), \(-\sqrt{5}\sqrt{5}\) are the intervals to be tested.

Test the intervals

Substitute a value from each interval into \(y''\) to find its sign. \(y''\) is positive for \(-\sqrt{5}\sqrt{5}\), and negative for \(x<-\sqrt{5}\) and \(0\sqrt{5}\), and concave downward on the intervals \(x<-\sqrt{5}\) and \(0<x<\sqrt{5}\).