Let f,g:RR be two real valued functions defined as f(x)=cases-|x+3|, & x

Question

Let $f,g:\mathbb{R}	o\mathbb{R}$ be two real valued functions defined as $f(x)=\begin{cases}-|x+3|, & x<0\ e^x, & x\ge0\end{cases}$ and $g(x)=\begin{cases}x^2+k_1x, & x<0\ 4x+k_2, & x\ge0\end{cases}$, where $k_1$ and $k_2$ are real constants. If $(g\circ f)$ is differentiable at $x=0$, then $(g\circ f)(-4)+(g\circ f)(4)$ is equal to:

Accepted Answer

For $x<0$ near $0$, $x+3>0$, so $f(x)=-(x+3)$, giving $f(x)\to-3$ as $x\to0^-$ (negative), so $g(f(x))=f(x)^2+k_1f(x)$ there. For $x\ge0$, $f(x)=e^x>0$, so $g(f(x))=4f(x)+k_2$. Continuity at $x=0$: $\lim_{x\to0^-}[f(x)^2+k_1f(x)]=9-3k_1$, and $\lim_{x\to0^+}[4e^x+k_2]=4+k_2$. Equate: $9-3k_1=4+k_2$ ...(a) Differentiability at $x=0$: For $x<0$: $g(f(x))=f(x)^2+k_1f(x)$ with $f(x)=-x-3,\ f'(x)=-1$, so $\dfrac{d}{dx}[g(f(x))]=f'(x)[2f(x)+k_1]=-1[-2x-6+k_1]$, which at $x=0$ gives $6-k_1$. For $x>0$: $\dfrac{d}{dx}[4e^x+k_2]=4e^x\to4$ at $x=0$. Equate: $6-k_1=4\implies k_1=2$. From (a): $9-6=4+k_2\implies k_2=-1$. Evaluate: $f(-4)=-|-4+3|=-1$ (negative branch): $(g\circ f)(-4)=(-1)^2+2(-1)=1-2=-1$. $f(4)=e^4$ (positive branch): $(g\circ f)(4)=4e^4+k_2=4e^4-1$. $$(g\circ f)(-4)+(g\circ f)(4)=-1+4e^4-1=4e^4-2=2(2e^4-1)$$

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