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  • Examine the differentiability of f, where f is defined by f(x)={x[x],(x-1)x

Examine the differentiability of f, where f is defined by

f(x)=\left\{\begin{array}{cc} x[x], & \text { if } 0 \leq x<2 \\ (x-1) x, & \text { if } 2 \leq x<3 \end{array}\right. \text { at } x=2.

Answers (1)

Given,

f(x)=\left\{\begin{array}{cc} x[x], & \text { if } 0 \leq x<2 \\ (x-1) x, & \text { if } 2 \leq x<3 \end{array}\right. \text { at } x=2  …(1)

We need to check whether f(x) is continuous and differentiable at x = 2

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)

Where h is a very small number very close to 0 (h→0)

And a function is said to be differentiable at x = c if it is continuous there and

Left hand derivative (LHD at x = c) = Right hand derivative (RHD at x = c) = f(c).

Mathematically we can represent it as-

\lim _{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c}=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c}

\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{c-h-c}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{c+h-c}

Finally, we can state that for a function to be differentiable at x = c

\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{-h}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}

Checking for the continuity:

Now according to above theory-

f(x) is continuous at x = 2 if -

\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2)

\therefore \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(2-\mathrm{h})

\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0}(2-\mathrm{h})[2-\mathrm{h}]_{\{\text {using equation } 1\}}

Note: As [.] represents greatest integer function which gives greatest integer less than the number inside [.].

E.g. [1.29] = 1; [-4.65] = -5 ; [9] = 9

[2-h] is just less than 2 say 1.9999 so [1.999] = 1

⇒ LHL = (2-0) ×1

∴ LHL = 2 …(2)

Similarly,

RHL = \lim _{h \rightarrow 0} f(2+h)

⇒ RHL = \lim _{h \rightarrow 0}(2+h-1)(2+h)_{\{\text {using equation } 1\}}

\therefore \mathrm{RHL}=(1+0)(2+0)=2 \ldots(3)

And, f(2) = (2-1)(2) = 2 …(4) {using equation 1}

From equation 2,3 and 4 we observe that:

\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2)=2

∴ f(x) is continuous at x = 2. So we will proceed now to check the differentiability.

Checking for the differentiability:

Now according to above theory-

f(x) is differentiable at x = 2 if -

\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}

\therefore LHD = \lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}

⇒ LHD = \lim _{h \rightarrow 0} \frac{(2-h)[2-h]-(2-1)(2)}{-h} \quad\{\text { using equation } 1\}

Note: As [.] represents greatest integer function which gives greatest integer less than the number inside [.].

E.g. [1.29] = 1 ; [-4.65] = -5 ; [9] = 9

 [2-h] is just less than 2 say 1.9999 so [1.999] = 1
 

⇒ LHD = \lim _{h \rightarrow 0} \frac{(2-h) \times 1-2}{-h}

⇒ LHD =  \lim _{h \rightarrow 0} \frac{-h}{-h}=\lim _{h \rightarrow 0} 1

\therefore \mathrm{LHD}=1 \underline{\ldots}(5)

Now,

RHD =  \lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}

⇒ RHD = \lim _{h \rightarrow 0} \frac{(2+h-1)(2+h)-(2-1)(2)}{h} \quad\{\text { using equation } 1\}

⇒ RHD = \lim _{h \rightarrow 0} \frac{(1+h)(2+h)-2}{h}=\lim _{h \rightarrow 0} \frac{2+h^{2}+3 h-2}{h}

∴ RHD = \lim _{h \rightarrow 0} \frac{h(h+3)}{h}=\lim _{h \rightarrow 0}(h+3)

\Rightarrow \mathrm{RHD}=0+3=3 \underline{\ldots}(6)

Clearly from equation 5 and 6, we can conclude that- 

(LHD at x=2) ≠ (RHD at x = 2)

∴ f(x) is not differentiable at x = 2

 

Posted by

infoexpert21

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