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

Show that the function $f (x) = |x| + |x - 1|$ , $x \in R$ , is continuous both at $x = 0$ and $x = 1$ .

Accepted Answer

Given,

$f (x) = |x| + |x - 1|$

A function is said to be continuous at $x = 0$ if

$x = 1$

Now,

$L . H . L = \lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h)$ $= \lim_{h \to 0} |0 - h| + |0 - h - 1|$

$= \lim_{h \to 0} |- h| + \lim_{h \to 0} |- h - 1|$ $= \lim_{h \to 0} h + \lim_{h \to 0} (h + 1)$ ${∵ |a . b| = |a| . |b| \Rightarrow |- h| = |- 1| . |h| = 1 . h = h also |- h - 1| = |- 1 (1 + h)| = |- 1| |1 + h| = 1 . |1 + h|}$

$= 0 + 0 + 1$ $= 1 . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

$R . H . L = \lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h)$

$= \lim_{h \to 0} |0 + h| + |0 + h - 1|$ $= \lim_{h \to 0} |h| + \lim_{h \to 0} |h - 1|$

$= \lim_{h \to 0} h + \lim_{h \to 0} - (h - 1) {∵ h - 1 < 0 \Rightarrow |h - 1| = - (h - 1)}$

$= 0 + 1$ $= 1 . . . . . . . . . . . . . . . . . . . . . (2)$

$f (0) = |0| + |0 - 1| = 1 . . . . . . . . . . . . . . . . . . . . . (3)$ By definition of function

Taking $(1), (2) and (3)$ equation

$∵ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0)$

Hence, the function $f (x)$ Is continuous at $x = 0$ .

At $x = 1$

Given,

$f (x) = |x| + |x - 1|$

A function is said to be continuous at $x = 1$ if

$\lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1)$

Now,

$L . H . L = \lim_{x \to 1^{-}} f (x) = \lim_{h \to 0} f (1 - h)$ $= \lim_{h \to 0} |1 - h| + |1 - h - 1|$

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

${∵ 1 - h > 0 \Rightarrow |1 - h| = 1 - h |- h| = |- 1| |h| = 1 . h = h}$

$= 1 + 0$ $= 1 . . . . . . . . . . . . . . . . . . . . . . . . . (4)$

$R . H . L = \lim_{x \to 1^{+}} f (x) = \lim_{h \to 0} f (1 + h)$

$= \lim_{h \to 0} |1 + h| + \lim_{h \to 0} h {∵ (1 + h) > 0 \Rightarrow |1 + h| = 1 + h}$

$= 1 + 0$ $= 1 . . . . . . . . . . . . . . . . . . . . . (5)$

$f (1) = |1| + |1 - 1| = 1 . . . . . . . . . . . . . . . . . . . . . (6)$ By definition of function

Taking $(4), (5) and (6)$ equation

$∵ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1)$

Hence, the function $f$ Is continuous at $x = 1$ .

Show that the function $f (x) = |x| + |x - 1|$ , $x \in R$ , is continuous both at $x = 0$ and $x = 1$ .

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

Show that the function fx=x+x-1, x∈R, is continuous both at x=0 and x=1.

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

Show that the function $f (x) = |x| + |x - 1|$ , $x \in R$ , is continuous both at $x = 0$ and $x = 1$ .