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Question: The function \[f\left( x \right)=\left\\{ \begin{aligned} & \left| x-3 \right|,\text{ }x\ge 1 \\...

The function f\left( x \right)=\left\\{ \begin{aligned} & \left| x-3 \right|,\text{ }x\ge 1 \\\ & \dfrac{{{x}^{2}}}{4}-\dfrac{3x}{2}+\dfrac{13}{4},\text{ }x<1 \\\ \end{aligned} \right. is
(a) continuous at x=1x=1
(b) differentiable at x=1x=1
(c) continuous at x=3x=3
(d) differentiable at x=3x=3

Explanation

Solution

Hint: If the value of limit of the function at a point x=ax=a is equal to the value of the function at x=ax=a , the function is said to be continuous at x=ax=a. A function is differentiable at x=ax=a , if the left-hand derivative of the function is equal to the right hand derivative of the function at x=ax=a.

Complete step-by-step answer:
At x=3x=3, f(x)=∣x−3∣f\left( x \right)=\left| x-3 \right|.
We know ∣x−a∣\left| x-a \right| is a function that is continuous everywhere but it is not differentiable at vertex i.e. at x=ax=a.
So , f(x)f\left( x \right)is continuous at x=3x=3 but not differentiable at x=3x=3.
Now , we will check if the function is continuous or differentiable at critical points of the function .
A function is differentiable at x=ax=a , if the left-hand derivative of the function is equal to the right hand derivative of the function at x=ax=a.
We know , the left hand derivative of f(x)f\left( x \right) at x=ax=a is given as L′=lim⁡h→0− f(a−h)−f(a)−h{{L}^{'}}=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( a-h \right)-f\left( a \right)}{-h} and the right hand derivative of f(x)f\left( x \right)at x=ax=a is given as R′=lim⁡h→0+ f(a+h)−f(a)h{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}.
We will consider the critical point x=1x=1. To determine differentiability of f(x)f\left( x \right) at x=1x=1, we will evaluate its left-hand derivative.
L′=lim⁡h→0− f(1−h)−f(1)−h{{L}^{'}}=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( 1-h \right)-f\left( 1 \right)}{-h}
=lim⁡h→0− [((1−h)24−3(1−h)2+134)−(124−3(1)2+134)−h]=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \dfrac{\left( \dfrac{{{\left( 1-h \right)}^{2}}}{4}-\dfrac{3\left( 1-h \right)}{2}+\dfrac{13}{4} \right)-\left( \dfrac{{{1}^{2}}}{4}-\dfrac{3\left( 1 \right)}{2}+\dfrac{13}{4} \right)}{-h} \right]
=lim⁡h→0− [h2−2h+14−3−3h2+134−14+32−134−h]=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \dfrac{\dfrac{{{h}^{2}}-2h+1}{4}-\dfrac{3-3h}{2}+\dfrac{13}{4}-\dfrac{1}{4}+\dfrac{3}{2}-\dfrac{13}{4}}{-h} \right]
=lim⁡h→0− [h2+4h−4h]=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \dfrac{{{h}^{2}}+4h}{-4h} \right]
=lim⁡h→0− [h+4−4]=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \dfrac{h+4}{-4} \right]
=−1=-1
Now , we will evaluate the right-hand derivative of f(x)f\left( x \right) at x=1x=1
R′=lim⁡h→0+ f(1+h)−f(1)h{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( 1+h \right)-f\left( 1 \right)}{h}
R′=lim⁡h→0+ ∣1+h−3∣−∣1−3∣h{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left| 1+h-3 \right|-\left| 1-3 \right|}{h}
Now, we know1+h<31+h<3,
So ∣1+h−3∣=∣h−2∣=2−h\left| 1+h-3 \right|=\left| h-2 \right|=2-h
So, R′=lim⁡h→0+ 2−h−2h{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{2-h-2}{h}
=−1=-1
Clearly, the left hand derivative is equal to the right hand derivative.
Hence f(x)f\left( x \right) is differentiable at x=1x=1.
Now, we also know that if a function is differentiable at a point then it must be continuous at that point.
So, f(x)f\left( x \right) is continuous at x=1x=1.
Hence , the function is differentiable and continuous at x=1x=1 and the function is continuous but not differentiable at x=3x=3.
Answer is (a) , (b) , (c)

Note: A function which is differentiable at a point is necessarily continuous at that point. But a function which is continuous at a point, may or may not be differentiable at that point.