First slide

Differentiability

Question

$The function f (x) = \{\begin{aligned} | x - 3 |, x \geq 1 \\ \frac{x^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}, x < 1 \end{aligned}$

Moderate

A

discontinuous at x = 1
B

not differentiable at x = 1
C

continuous and differentiable at x = 1
D

continuous at x = 1 but not derivable at x = 1 .

Solution

$\begin{array}{l} f' (1^{-}) = \lim_{x \to 1^{-}} \frac{f (x) - f (1)}{x - 1} \\ = \lim_{x \to 1} \frac{\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{13}{4} - 2}{x - 1} \\ = \lim_{x \to 1} \frac{x^{2} - 6 x + 5}{4 (x - 1)} \\ = \lim_{x \to 1} \frac{(x - 5) (x - 1)}{4 (x - 1)} = \lim_{x \to 1} \frac{(x - 5)}{4} = - 1 \\ f' (1^{+}) = \lim_{x \to 1^{+}} \frac{f (x) - f (1)}{x - 1} \lim_{x \to 1} \frac{3 - x - 2}{x - 1} = - 1 \\ ∴ f' (1) = - 1 \\ ∴ f (x) is derivable and hence continuous also at x = 1 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App