Question

Hello, I am unsure how to get to the answer, can you include a sign chart as well for the second derivative thank you.

Answer 1

We want to find the intervals of concavity of the function;

$y=-x^4+8x^2-4$

We start by taking the second derivatives;

$\begin{gathered} y^{\prime}=-4x^3+16x \\ y^{\prime}^{\prime}=-12x^2+16 \end{gathered}$

When a function is concave up, the second derivative is positive, thus; we seek the intervals where;

$\begin{gathered} -12x^2+16>0 \\ -12x^2>-16 \\ x^2<\frac{16}{12} \\ x^2<\frac{4}{3} \\ x<\pm\sqrt{\frac{4}{3}} \end{gathered}$

Let's find the points of inflection, this is where the second derivative is zero;

$\begin{gathered} -12x^2+16=0 \\ -12x^2=-16 \\ x^2=\frac{16}{12}=\frac{4}{3} \\ x=\pm\frac{4}{3} \end{gathered}$

The y values will be;

$\begin{gathered} -(\frac{4}{3})^4+8(\frac{4}{3})^2-4=\frac{44}{9} \\ -(-\frac{4}{3})^4+8(-\frac{4}{3})^2-4=\frac{44}{9} \end{gathered}$

Thus, the answers are;

$\begin{gathered} Concave\text{ }up:(-\sqrt{\frac{4}{3}},\sqrt{\frac{4}{3}}) \\ Inflection\text{ }points:(-\sqrt{\frac{4}{3}},\frac{44}{9}),(\sqrt{\frac{4}{3}},\frac{44}{9}) \end{gathered}$

Thus the answer is option D;

This is a sign chart for the second derivative;