Let f(x) be a real valued function, then its * Left Hand Derivative (L.H.D.) : Lf'(a)= _ h 0 f(a-h)-f(a) -h * Right Hand Derivative (R.H.D.) : Rf'(a)= _ h 0 f(a+h)-f(a) h Also, a function f(x) is said to be differentiable at x = a if its L.H.D. and R.H.D. at x = a exist and are equal. For the function f(x)= |x-3|,x 1 x^2 4 - 3x 2 +13 4 ,x<1 , answer the following questions. 1. R.H.D. of f(x) at x = 1 is: 1. 1 2. -1 3. 0 4. 2 2. L.H.D. of f(x) at x = 1 is: 1. 1 2. -1 3. 0 4. 2 3. f(x) is non-differentiable at: 1. x = 1 2. x = 2 3. x = 3 4. x = 4 4. Find the value of f'(2). 1. 1 2. 2 3. 3 4. -1 5. The value of f'(-1) is: 1. 2 2. 1 3. -2 4. -1

We have, f(x)= x-3&,x 3 3-x&,1 <3 x^2 4 - 3x 2 +13 4 &,x<1 1. (b) -1 Solution: Rf'(1)= _ h 0 f(1+h)-f(1) h _ h 0 3-(1+h)-2 h = _ h 0 - h h =-1 2. (b) -1 Solution: Lf'(1)= _ h 0 f(1-h)-f(1) -h = _ h 0 -1 h [ (1-h)^2 4 - 3(1-h) 2 +13 4 -2 ] = _ h 0 -1 h ( 1+h^2-2h-6+6h+13-8 -4h ) = _ h 0 ( h^2+4h -4h )=-1 3. (c) x = 3 Solution: Since, R.H.D. at x = 3 is 1 and L.H.D. at x = 3 is - 1 f(x) is non-differentiable at x = 3. 4. (d) -1 5. (c) -2 Solution: From above, we have f'(x)= x 2 -3 2 ,x<1 '(-1)=-1 2 -3 2 =-2

Let f(x) be a real valued function, then its * Left Hand Derivati…

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Let f(x) be a real valued function, then its

Left Hand Derivative (L.H.D.) : $\text{Lf}'(\text{a})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{a}-\text{h})-\text{f}(\text{a})}{-\text{h}}$
Right Hand Derivative (R.H.D.) : $\text{Rf}'(\text{a})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{a}+\text{h})-\text{f}(\text{a})}{\text{h}}$

Also, a function f(x) is said to be differentiable at x = a if its L.H.D. and R.H.D. at x = a exist and are equal.
For the function $\text{f}(\text{x})=\begin{cases}|\text{x}-3|,\text{x}\geq1\\\\\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4},\text{x}<1\end{cases},$ answer the following questions.

R.H.D. of f(x) at x = 1 is:

1
-1
0
2

L.H.D. of f(x) at x = 1 is:

1
-1
0
2

f(x) is non-differentiable at:

x = 1
x = 2
x = 3
x = 4

Find the value of f'(2).

1
2
3
-1

The value of f'(-1) is:

2
1
-2
-1

✓

Answer

We have, $\text{f}(\text{x})=\begin{cases}\text{x}-3&,\text{x}\geq3\\3-\text{x}&,1\leq\text{x}<3\\\\\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4}&,\text{x}<1\end{cases}$

(b) -1

Solution:
$\text{Rf}'(1)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}$
$\lim\limits_{\text{h}\rightarrow0}\frac{3-(1+\text{h})-2}{\text{h}}=\lim\limits_{\text{h}\rightarrow0}-\frac{\text{h}}{\text{h}}=-1$

(b) -1

Solution:
$\text{Lf}'(1)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\text{h}}\Big[\frac{(1-\text{h})^2}{4}-\frac{3(1-\text{h})}{2}+\frac{13}{4}-2\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\text{h}}\Big(\frac{1+\text{h}^2-2\text{h}-6+6\text{h}+13-8}{-4\text{h}}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{h}^2+4\text{h}}{-4\text{h}}\Big)=-1$

Solution:
Since, R.H.D. at x = 3 is 1
and L.H.D. at x = 3 is - 1
$\therefore$ f(x) is non-differentiable at x = 3.

(d) -1

Solution:
From above, we have
$\text{f}'(\text{x})=\frac{\text{x}}{2}-\frac{3}{2},\text{x}<1$
$\therefore\text{f}'(-1)=\frac{-1}{2}-\frac{3}{2}=-2$

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