The function f(x)= cc |x-3|, & x 1 x^2 4 -3 x 2 +13 4 , & x<1 . is

(d) : |x-3| is continuous at x=3, but not differentiable. f (1^ - )=f (1^ + )=f(1)=2 f^ (1^ - )=f^ (1^ + )=-1=f^ (1) f(x) is continuous and differentiable at x=1.

The function f(x)= cc |x-3|, & x 1 x^2 4 -3 x 2 +13 4 , & x<1 . i…

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Question

The function $f(x)=\left\{\begin{array}{cc}|x-3|, & x \geq 1 \\ \frac{x^2}{4}-\frac{3 x}{2}+\frac{13}{4}, & x<1\end{array}\right.$ is

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Answer

$(d) : |x-3|$ is continuous at $x=3,$ but not differentiable.
$f\left(1^{-}\right)=f\left(1^{+}\right)=f(1)=2$
$f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right)=-1=f^{\prime}(1)$
$\therefore f(x)$ is continuous and differentiable at $x=1$.

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