Name: Differentiation of Exponential Functions
Uploaded: 2019-07-19
Duration: 12 min 3 s
Description: To know how to find the derivative of exponential functions or multiple functions containing the exponential term. Let's watch this video and explore!

Question

If the function

f (x) = |x - 3| + |x - 4|

, then show that

f

is not differentiable at

x = 3

and

x = 4

.

Accepted Answer

Given function is $f (x) = |x - 3| + |x - 4|$ ,

We have to show that $f$ is not differentiable at $x = 3$ and $x = 4$ .

The function $f (x)$ can be written as: $f (x) = \{\begin{matrix} - (x - 3) - (x - 4); x < 3 \\ (x - 3) - (x - 4); 3 \leq x < 4 \\ (x - 3) + (x - 4); x \geq 4 \end{matrix}$

$∴ f (x) = \{\begin{matrix} - 2 x + 7; x < 3 \\ 1; 3 \leq x < 4 \\ 2 x - 7; x \geq 4 \end{matrix}$

Differentiability at $x = 3$ :

$L f' (3) = \lim_{h \to 0} \frac{f (3 - h) - f (3)}{- h} = \lim_{h \to 0} \frac{[- 2 (3 - h) + 7] - 1}{- h} = \lim_{h \to 0} \frac{[- 2 \times 3 + 2 \times h + 7] - 1}{- h}$

$= \lim_{h \to 0} \frac{- 6 + 2 h + 7 - 1}{- h} = \lim_{h \to 0} \frac{2 h}{- h} = \lim_{h \to 0} (- 2) = - 2$

$R f' (3) = \lim_{h \to 0} \frac{f (3 + h) - f (3)}{h} = \lim_{h \to 0} \frac{1 - 1}{h} = \lim_{h \to 0} \frac{0}{h} = 0$

$∴ L f' (3) \neq R f' (3)$

Thus, $f$ is not differentiable at $x = 3$ .

Now, differentiability at $x = 4$ :

$L f' (4) = \lim_{h \to 0} \frac{f (4 - h) - f (4)}{- h} = \lim_{h \to 0} \frac{1 - [2 \times 4 - 7]}{- h} = \lim_{h \to 0} \frac{1 - (8 - 7)}{- h} = \lim_{h \to 0} \frac{1 - 1}{- h} = \lim_{h \to 0} \frac{0}{- h} = 0$

$R f' (4) = \lim_{h \to 0} \frac{f (4 + h) - f (4)}{h} = \lim_{h \to 0} \frac{[2 (4 + h) - 7] - (2 \times 4 - 7)}{h}$

$= \lim_{h \to 0} \frac{(2 \times 4 + 2 \times h - 7) - (8 - 7)}{h} = \lim_{h \to 0} \frac{(8 + 2 h - 7) - 1}{h} = \lim_{h \to 0} \frac{1 + 2 h - 1}{h} = \lim_{h \to 0} \frac{2 h}{h} = \lim_{h \to 0} 2 = 2$

$∴ L f' (4) \neq R f' (4)$

Thus, $f (x)$ is not differentiable at $x = 4$ .

Hence, $f$ is not differentiable at $x = 3$ and $x = 4$ .

If the function $f (x) = |x - 3| + |x - 4|$ , then show that $f$ is not differentiable at $x = 3$ and $x = 4$ .

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

If the function fx=x-3+x-4, then show that f is not differentiable at x=3 and x=4.

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

If the function $f (x) = |x - 3| + |x - 4|$ , then show that $f$ is not differentiable at $x = 3$ and $x = 4$ .