If the function f(x)=casesk_1(x-)^2-1, & x_2 x, & xcases is twice differentiable, then th…

Question

If the function $f(x)=\begin{cases}k_1(x-\pi)^2-1, & x\le\pi\k_2\cos x, & x>\pi\end{cases}$ is twice differentiable, then the ordered pair $(k_1,k_2)$ is equal to:

Accepted Answer

Continuity at $\pi$: $-1=k_2\cos\pi=-k_2\Rightarrow k_2=1$.

$f'(x)=\begin{cases}2k_1(x-\pi), & x\le\pi\-k_2\sin x, & x>\pi\end{cases}$; both sides $	o0$ at $\pi$, consistent for any $k_1$.

$f''(x)=\begin{cases}2k_1, & x\le\pi\-k_2\cos x, & x>\pi\end{cases}$; matching: $2k_1=k_2=1\Rightarrow k_1=\dfrac12$.

So $(k_1,k_2)=\left(\dfrac12,1ight)$.

Solutions · 1