Lagrange's theorem on a piecewise exponential function

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Let $f(x)=\begin{cases}a x+b & -1\le x\le 0\\[2pt] e^{x} & 0<x\le 1\end{cases}\qquad a,b\in\mathbb{R}.$ (a) Find $a$ and $b$ for which the hypotheses of Lagrange’s theorem hold on $[-1,1]$ (continuity and differentiability at $x=0$). (b) With those values, apply Lagrange’s theorem on $[-1,1]$: compute the global average rate of change and all $c\in(-1,1)$ with $f'(c)$ equal to it. How many are there?

Solution

(a) Conditions on the parameters. The only critical point is the junction $x=0$.

Continuity at $x=0$: left-hand limit $a\cdot 0+b=b$, right-hand value $e^0=1$. Hence $b=1$.

Differentiability at $x=0$: left-hand derivative $f'=a$, right-hand derivative $f'=e^x\big|_{0}=1$. Hence $a=1$.

(b) Applying the theorem. With $a=1$, $b=1$: $f(-1)=1\cdot(-1)+1=0,\qquad f(1)=e^1=e,$ so the average rate of change is $\frac{f(1)-f(-1)}{1-(-1)}=\frac{e-0}{2}=\frac{e}{2}\approx 1.359.$

On $(-1,0)$: $f'=1\ne\dfrac{e}{2}$, no point $c$.

On $(0,1)$: $f'=e^x=\dfrac{e}{2}\Rightarrow x=\ln\dfrac{e}{2}=1-\ln 2\approx 0.307\in(0,1)$ ✓

There is therefore exactly one point: $c=1-\ln 2\approx 0.3069.$

$\boxed{a=1,\qquad b=1,\qquad c=1-\ln 2\approx 0.3069\ (\text{unique})}$