The function f(x) = begin{cases} |x - 3| & x | vedclass

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(D) First,we analyze the continuity and differentiability at $x = 1$ and $x = 3$.For $x = 1$:Left-hand limit: $\lim_{x 	o 1^-} f(x) = \frac{1^2}{4} -

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All of the above

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(D) First,we analyze the continuity and differentiability at $x = 1$ and $x = 3$.For $x = 1$:Left-hand limit: $\lim_{x 	o 1^-} f(x) = \frac{1^2}{4} - \frac{3(1)}{2} + \frac{13}{4} = \frac{1 - 6 + 13}{4} = 2$.Right-hand limit: $\lim_{x 	o 1^+} f(x) = |1 - 3| = 2$.Since $f(1) = 2$,the function is continuous at $x = 1$.Now check differentiability at $x = 1$:$f'(1^+) = \frac{d}{dx}(3 - x) = -1$ (for $1 \leqslant x $f'(1^-) = \frac{d}{dx}(\frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}) = \frac{2x}{4} - \frac{3}{2} = \frac{1}{2} - \frac{3}{2} = -1$.Since $f'(1^+) = f'(1^-)$,the function is differentiable at $x = 1$.For $x = 3$:The function $f(x) = |x - 3|$ is continuous at $x = 3$ because $\lim_{x 	o 3} |x - 3| = 0 = f(3)$.Thus,all statements are correct.

The function $f(x) = \begin{cases} |x - 3| & x \geqslant 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4} & x < 1 \end{cases}$ is :

Similar Questions

Which of the following statements is true for the function $f(x) = \begin{cases} \sqrt{x} & x \ge 1 \\ x^3 & 0 \le x < 1 \\ \frac{x^3}{3} - 4x & x < 0 \end{cases}$

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