Question

Let $f\left( x \right) = {x^{\dfrac{3}{2}}} - \sqrt {{x^3} + {x^2}} $, then
1) LHD at $x = 0$ but RHD at $x = 0$ does not exist
2) $f\left( x \right)$ is differentiable at x=0
3) RHD at $x = 0$ exist but LHD at $x = 0$ does not exist
4) $f\left( x \right)$ is differentiable and continuous at $x = 0$

Answer 1

Hint: We are to find the RHD and LHD of the given function by using the fundamental theorem of differentiation, that is,
For $x > a$, RHD $ = \mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f\left( {a + h} \right) - f(a)}}{h}$
For $x < a$, LHD $ = \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f\left( {a - h} \right) - f\left( a \right)}}{{ - h}}$
We are to solve these equations, to find the right hand derivative and left hand derivative and observe which of the conditions get satisfied.

Complete step by step answer:
The given function is, $f\left( x \right) = {x^{\dfrac{3}{2}}} - \sqrt {{x^3} + {x^2}} $
Now, we are to find the RHD and LHD at $x = 0$.
Therefore, for $x > 0$, RHD=$\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f\left( {0 + h} \right) - f(0)}}{h}$
Now, evaluating the function, we get,
RHD$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{\left\{ {{{\left( {0 + h} \right)}^{\dfrac{3}{2}}} - \sqrt {{{\left( {0 + h} \right)}^3} + {{\left( {0 + h} \right)}^2}} } \right\} - \left\{ {{{\left( 0 \right)}^{\dfrac{3}{2}}} - \sqrt {{{\left( 0 \right)}^3} + {{\left( 0 \right)}^2}} } \right\}}}{h}$
$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{\left( {{h^{\dfrac{3}{2}}} - \sqrt {{h^3} + {h^2}} } \right) - 0}}{h}$
$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{{h^{\dfrac{3}{2}}} - h\sqrt {1 + h} }}{h}$
Taking $h$ common from numerator, we get,
$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{h\left( {{h^{\dfrac{1}{2}}} - \sqrt {1 + h} } \right)}}{h}$
Cancelling $h$from numerator and denominator, we get,
$ = \mathop {\lim }\limits_{h \to {0^ + }} \left( {{h^{\dfrac{1}{2}}} - \sqrt {1 + h} } \right)$
Now, evaluating the limit, we get,
$ = 0 - \sqrt {1 + 0} $
$ = - 1$
Therefore, RHD exists at $x = 0$ and it’s value is $ - 1$.
Now, for $x < 0$, LHD=$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f\left( {0 - h} \right) - f(0)}}{{ - h}}$
Now, evaluating the function, we get,
LHD$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{\left\{ {{{\left( {0 - h} \right)}^{\dfrac{3}{2}}} - \sqrt {{{\left( {0 - h} \right)}^3} + {{\left( {0 - h} \right)}^2}} } \right\} - \left\{ {{{\left( 0 \right)}^{\dfrac{3}{2}}} - \sqrt {{{\left( 0 \right)}^3} + {{\left( 0 \right)}^2}} } \right\}}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{\left\{ {{{\left( { - h} \right)}^{\dfrac{3}{2}}} - \sqrt {{{\left( { - h} \right)}^3} + {{\left( { - h} \right)}^2}} } \right\} - \left\{ 0 \right\}}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{ - {h^{\dfrac{3}{2}}} - \sqrt { - {h^3} + {h^2}} }}{{ - h}}$
Taking $h$ common from the root, we get,
$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{ - {h^{\dfrac{3}{2}}} - h\sqrt { - h + 1} }}{{ - h}}$
Now, taking $ - h$ common from the numerator, we get,
$ = \mathop {\lim }\limits_{h \to {0^ - }} {h^{\dfrac{1}{2}}} + \sqrt {1 - h} $
Here, $h \to {0^ - }$, i.e., $h < 0$, so $\sqrt h $ will not exist, as it can’t be negative inside the root.
So, LHD doesn’t exist.
So, RHD exists at $x = 0$, but LHD does not exist at $x = 0$.
Therefore, the option (3) is correct.

Note:
> If both the RHD and LHD of a function exists, then the function is said to be differentiable at the given point. But if any one of the LHD or RHD doesn’t exist, then the function is not differentiable at that given point. This fundamental theorem is the basis of the origin of the formulas of differentiation.
> If a function is differentiable, then that function must be continuous but the converse does not hold. That means, if a function is continuous, it might not always be differentiable.