If the function f: R rightarrow R, defined by | vedclass

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(C) We check the continuity at $x = \frac{5}{3}$:Left-hand limit: $\lim_{x 	o \frac{5}{3}^-} f(x) = 5 - 3(\frac{5}{3}) = 5 - 5 = 0$.Right-hand limit:

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differentiable at $x = 2$

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(C) We check the continuity at $x = \frac{5}{3}$:Left-hand limit: $\lim_{x 	o \frac{5}{3}^-} f(x) = 5 - 3(\frac{5}{3}) = 5 - 5 = 0$.Right-hand limit: $\lim_{x 	o \frac{5}{3}^+} f(x) = (\frac{5}{3})^2 - 3(\frac{5}{3}) + 20 = \frac{25}{9} - 5 + 20 = \frac{25}{9} + 15 = \frac{25 + 135}{9} = \frac{160}{9}$.Since $\lim_{x 	o \frac{5}{3}^-} f(x) 
eq \lim_{x 	o \frac{5}{3}^+} f(x)$, the function is discontinuous at $x = \frac{5}{3}$.Now check differentiability at $x = 2$. Since $2 > \frac{5}{3}$, the function is defined by $f(x) = x^2 - 3x + 20$ in the neighborhood of $x = 2$.This is a polynomial function, which is differentiable everywhere in its domain. Thus, $f(x)$ is differentiable at $x = 2$.

If the function $f: R \rightarrow R$, defined by $f(x) = \begin{cases} 5-3x, & \text{if } x \leq \frac{5}{3} \\ x^2-3x+20, & \text{if } x > \frac{5}{3} \end{cases}$, then $f$ is

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