?Let Find the values of and that make differentiable everywhere

Step 1 of 2

Given function is

\(\begin{aligned}
f(x) & =x^{2}, \quad x \leq 2 \\
& =m x+b, x>2
\end{aligned}\)

To find the values of m and b that makes f differentiable everywhere.

We will discuss the differentiability of the function at x = 2.

For \(x \leq 2\),

\(\begin{aligned}
f^{\prime}(x) & =2 x \\
f^{\prime}(2) & =4
\end{aligned}\)

Thus, \(L f^{\prime}(2)=4\)

For x > 2,

\(\begin{array}{l}
f^{\prime}(x)=m \\
f^{\prime}(2)=m
\end{array}\)

Thus, \(R f^{\prime}(2)=m\)

For the function f to be continuous at \(x=2\), we must have

\(L f^{\prime}(2)=R f^{\prime}(2)\)

It follows

m = 4