Name: Derivative of Functions in Parametric Form
Uploaded: 2019-07-19
Duration: 8 min 37 s
Description: Learn everything you'll need to know about derivatives of functions in parametric forms with this quick and easy lesson that'll cover the concept comprehensi...

Question

Show that

f (x) = | x - 5 |

is continuous but not differentiable at

x = 5

.

Accepted Answer

Given,

$f (x) = | x - 5 |$

$\Rightarrow f (x) = \{\begin{array}{r} - (x - 5) if x - 5 < 0 or x < 5 \\ x - 5 if x - 5 > 0 or x > 5 \end{array}$

For continuity at $x = 5$

Left hand limit:

$L.H.L. \lim_{x \to 5^{-}} f (x) = \lim_{x \to 5^{-}} \{- (x - 5)\}$

$= \lim_{h \to 0} - (5 - h - 5) = \lim_{h \to 0} h = 0$

Right hand limit:

$R.H.L. \lim_{x \to 5^{+}} f (x) = \lim_{x \to 5^{+}} (x - 5)$

$= \lim_{h \to 0} (5 + h - 5) = \lim_{h \to 0} h = 0$

Value of the function at $x = 5$ : $f (5) = 0$

$L . H . L . = R . H . L . = Value of f {(x)}_{x = 5}$

So, $f (x)$ is continuous at $x = 5$ .

Now, for differentiability,

We know that, a function $f$ is differentiable at a point $x = a$ if

${Lf}^{'} (a) = {Rf}^{'} (a)$ where,

$L f^{'} (a) = \lim_{h \to 0} \frac{f (a - h) - f (a)}{- h}$ and $R f^{'} (a) = \lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$

$L f^{'} (5) = \lim_{h \to 0} \frac{f (5 - h) - f (5)}{- h}$

$= \lim_{h \to 0} \frac{- (5 - h - 5) - (5 - 5)}{- h} = \lim_{h \to 0} \frac{h}{- h} = - 1$

$R f^{'} (5) = \lim_{h \to 0} \frac{f (5 + h) - f (5)}{h}$

$= \lim_{h \to 0} \frac{(5 + h - 5) - (5 - 5)}{h} = \lim_{h \to 0} \frac{h - 0}{h} = 1$

Thus, $L f^{'} (5) \neq R f^{'} (5)$

Therefore, $f (x)$ is not differentiable at $x = 5$ .

Show that $f (x) = | x - 5 |$ is continuous but not differentiable at $x = 5$ .

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

Show that f(x)=|x-5| is continuous but not differentiable at x=5.

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

Show that $f (x) = | x - 5 |$ is continuous but not differentiable at $x = 5$ .