Lagrange's theorem on a piecewise function with parameters

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Let $f(x)=\begin{cases}h\,x^3 & 0\le x\le 1\\[2pt] x^2-kx+k & 1<x\le 2\end{cases}$ (a) Find the values of $h$ and $k$ for which the hypotheses of Lagrange’s theorem hold on $[0,2]$. (b) With those values, apply Lagrange’s theorem and find a point $c\in(0,2)$.

Solution

(a) Conditions on the parameters. The hypotheses require continuity on $[0,2]$ and differentiability on $(0,2)$; the only critical point is the junction $x=1$.

Continuity at $x=1$: the left-hand limit is $h\cdot 1^3=h$, the right-hand one $1^2-k\cdot 1+k=1$. Hence $h=1.$

Differentiability at $x=1$: the derivatives of the two branches are $3h x^2$ and $2x-k$. At $x=1$: $3h=2-k \implies 3=2-k \implies k=-1.$

(b) Applying the theorem. With $h=1$, $k=-1$: $f(0)=0,\qquad f(2)=2^2-(-1)\cdot 2+(-1)=4+2-1=5,$ so the average rate of change is $\dfrac{f(2)-f(0)}{2-0}=\dfrac52$.

Left branch ($f'=3x^2$): $3x^2=\dfrac52 \Rightarrow x=\sqrt{5/6}\approx 0.913\in(0,1)$ ✓

Right branch ($f'=2x+1$): $2x+1=\dfrac52 \Rightarrow x=0.75\notin(1,2)$, discarded.

Therefore $c=\sqrt{\tfrac56}\approx 0.9129.$

$\boxed{h=1,\qquad k=-1,\qquad c=\sqrt{\tfrac56}\approx 0.9129}$