If a function f(x) defined by f(x)=casesae^x+be^-x, & -1 x

Question

If a function $f(x)$ defined by $f(x)=\begin{cases}ae^x+be^{-x}, & -1\le x<1\cx^2, & 1\le x\le3\ax^2+2cx, & 3<x\le4\end{cases}$ is continuous for some $a,b,c\in\mathbb{R}$ and $f'(0)+f'(2)=e$, then the value of $a$ is:

Accepted Answer

Continuity at $x=1$: $ae+be^{-1}=c$ ... (i). Continuity at $x=3$: $9c=9a+6c\Rightarrow c=3a$ ... (ii).

From (i),(ii): $b=ae(3-e)$ ... (iii).

$f'(x)=\begin{cases}ae^x-be^{-x}, & -1<x<1\2cx, & 1<x<3\2ax+2c, & 3<x<4\end{cases}$

$f'(0)=a-b$, $f'(2)=4c$. Given $a-b+4c=e$. Substituting (ii),(iii): $a-3ae+ae^2+12a=e\Rightarrow a(13-3e+e^2)=e\Rightarrow a=\dfrac{e}{e^2-3e+13}$.

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