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For $f(x)$ to be differentiable at $x=1$, it must first be continuous at $x=1$. We need to check if $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 1) = (1)^2 + 1 = 2$
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3-x) = 3 - 1 = 2$
$f(1) = (1)^2 + 1 = 2$
Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 2$, the function is continuous at $x=1$.
Now, we need to check if the left-hand derivative (LHD) is equal to the right-hand derivative (RHD) at $x=1$.
LHD = $\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{((1+h)^2 + 1) - 2}{h} = \lim_{h \to 0^-} \frac{1 + 2h + h^2 + 1 - 2}{h} = \lim_{h \to 0^-} \frac{2h + h^2}{h} = \lim_{h \to 0^-} (2 + h) = 2$
RHD = $\lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{(3 - (1+h)) - 2}{h} = \lim_{h \to 0^+} \frac{3 - 1 - h - 2}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1$
Since LHD = $2$ and RHD = $-1$, LHD $\neq$ RHD. Therefore, the function is not differentiable at $x=1$.
Final Answer: Not differentiable at x=1
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