Question

Find the derivative of the absolute value function and piecewise defined function

$f\left(x\right)=\left\{\begin{array}{ll}3x+1& ifx\le 1\\ x^3& ifx>1\end{array}\right.$

Short answer

The derivative of the piecewise function is$f\text{'}\left(x\right)=\left\{\begin{array}{ll}3& ifx\ge 1\\ 3x^2& ifx<1\end{array}\right.$

Step-by-step solution

Given information

We are given A function$f\left(x\right)=\left\{\begin{array}{ll}3x+1& ifx\le 1\\ x^3& ifx>1\end{array}\right.$

Find the derivative

We have

When $x\ge 1$

f(x)=3x+1

Then $f\text{'}\left(x\right)=3$

When x<1

$$\begin{gathered} f\left(x\right)=x^3 \\ f\text{'}\left(x\right)=3x^2 \end{gathered}$$

Now we have to check the derivative at 1

As the value of the left hand limit is equal to the right hand limit the function is continuous

Now,

We calculate the value of both the derivatives at 1

We get it as 3 for both the cases

Hence the function is differentiable at x=1

Hence

$f\text{'}\left(x\right)=\left\{\begin{array}{ll}3& ifx\ge 1\\ 3x^2& ifx<1\end{array}\right.$