Chapter 2.7, Problem 40E

To determine

To graph: The function $f\left(x\right)={\left(x^2-1\right)}^{\frac{2}{3}}$, and conclude the differentiability of the points $\left(1,0\right),\left(0,1\right)$ and $\left(-1,0\right)$.

Answer to Problem 40E

The function $f\left(x\right)$ is not differentiable at $\left(-1,0\right)$ and $\left(1,0\right)$ but differentiable at $\left(0,1\right)$.

Explanation of Solution

Given:

The given function is $f\left(x\right)={\left(x^2-1\right)}^{\frac{2}{3}}$.

Graph:

Use the online graphing calculator to draw the graph of the function $f\left(x\right)={\left(x^2-1\right)}^{\frac{2}{3}}$ as shown below in Figure 1.

Use the online graphing calculator to zoom toward the point $\left(1,0\right)$ as shown below in Figure 2.

From Figure 2, it is observed that when zoomed in the graph toward the point $\left(1,0\right)$, the curve has high sharpness at $\left(1,0\right)$. So, there is no unique tangent line at $\left(1,0\right)$.

Thus, $\left(1,0\right)$ is a corner point.

Hence, the given function $f\left(x\right)={\left(x^2-1\right)}^{\frac{2}{3}}$ is not differentiable at $\left(1,0\right)$.

Use the online graphing calculator to zoom toward the point $\left(-1,0\right)$ as shown below in Figure 3.

From Figure 3, it is observed that when zoomed in the graph toward the point $\left(-1,0\right)$, the curve has high sharpness at $\left(-1,0\right)$. So, there is no unique tangent line at $\left(-1,0\right)$.

Thus, $\left(-1,0\right)$ is a corner point.

Hence, the function $f\left(x\right)={\left(x^2-1\right)}^{\frac{2}{3}}$ is not differentiable at $\left(-1,0\right)$.

Use the online graphing calculator to zoom toward the point $\left(0,1\right)$ as shown below in Figure 4.

From Figure 4, it is observed that when zoomed in the graph toward the point $\left(0,1\right)$, the curve and the tangent are indistinguishable.

It has a unique tangent.

Hence, the function $f\left(x\right)={\left(x^2-1\right)}^{\frac{2}{3}}$ is differentiable at $\left(0,1\right)$.

Conclusion:

By zooming the graph toward the point $\left(-1,0\right)$ and $\left(1,0\right)$, the graph has no unique tangent line while zooming the graph toward the point $\left(0,1\right)$, the graph has unique tangent line.

Thus, the function $f\left(x\right)={\left(x^2-1\right)}^{\frac{2}{3}}$ is not differentiable at $\left(-1,0\right)$ and $\left(1,0\right)$ but differentiable at $\left(0,1\right)$.