Question

In Exercises \(31-36,\) find all values of \(x\) for which the function is differentiable. $$h(x)=\sqrt[3]{3 x-6}+5$$

Short answer

For all values of \(x\) except \(x=2\), the function \(h(x)=\sqrt[3]{3x-6}+5\) is differentiable.

Step-by-step solution

- Understand the function

The given function is \(h(x)=\sqrt[3]{3x-6}+5\).

- Compute the derivative

To find the values of \(x\) for which the function is differentiable, first, compute the derivative of the function using the power rule of derivative which states that the derivative of \(x^n\), where \(n\) is any real number, is given by \(nx^{n-1}\). Here, the power \(n\) is \(\frac{1}{3}\). After applying the chain rule, the derivative of \(h(x)\) is \(\frac{1}{3}(3x-6)^{-2/3}\).

- Find the values of x where function is differentiable

The function is differentiable at a point if it has a derivative at that point. From Step 2, there is one value at which the derivative does not exist because of division by zero, that is, when \(3x-6=0 \Rightarrow x=2\). For all values of \(x\) other than \(x=2\), the function is differentiable.