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**Hint:**We will start by using the conditions to be satisfied by a function which is differentiable at any point, when it is continuous at that point:

$\begin{align}

& f'\left( {{a}^{+}} \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}=finite \\

& f'\left( {{a}^{-}} \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( a-h \right)-f\left( a \right)}{h}=finite \\

\end{align}$ , where $h \to positive$

$f'\left( {{a}^{+}} \right)=f'\left( {{a}^{-}} \right)=f'\left( a \right)=finite~~value$ , when the function is differentiable at point $a$.

**Complete step-by-step solution**Since we have conditions given as $f'\left( x \right)>0$, we will apply this and form two equations for a positive and negative value of h. Then, we will use $f'\left( a \right)=0$ in the equation formed for the positive value of h to conclude the theorem.

Let $f$ is a function, which is continuous and differentiable in the neighborhood of point $a$ and at point $a$ also.

As we know that a function is differentiable at any point, when it is continuous at that point and following condition satisfies –

$\begin{align}

& f'\left( {{a}^{+}} \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}=finite \\

& f'\left( {{a}^{-}} \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( a-h \right)-f\left( a \right)}{h}=finite \\

\end{align}$ , where $h \to positive$

And when $f'\left( {{a}^{+}} \right)=f'\left( {{a}^{-}} \right)=f'\left( a \right)=finite~~value$, then function is differentiable at point $a$.

According to definition of differentiability,

\[f'\left( a \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}\] , where $h$ may be positive or negative both.

Now in theorem point $a$ is removed, means we cannot put value of $x$ as $a$ in given function $f\left( x \right)$.

So, let the curve of given function be $f\left( x \right)$. Given $f'\left( a \right)=0$ and $f'\left( x \right)>0$.

We know that, \[f'\left( a \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}>0\] .

If we take $h$ as positive,

\[\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}>0\]

\[\Rightarrow f\left( x+h \right)-f\left( x \right)>0\]

\[\Rightarrow f\left( x+h \right)>f\left( x \right).......(i)\]

It means that if we increase the value of $x$, then value of function is also increasing.

Now, if we take $h$ as negative,

\[\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}>0\]

\[\Rightarrow f\left( x+h \right)-f\left( x \right)< 0\]

\[\Rightarrow f\left( x+h \right)< f\left( x \right).......(ii)\]

Here, $h$ is negative, means that if we decrease the value of $x$, then the value of function is also decreasing.

So, according to these relations, curve of given $f\left( x \right)$ will be –

Here, $f'\left( a \right)=0$

In equation (i), where $h>0$, put $x=a$ .

\[\begin{align}

& \Rightarrow f\left( x+h \right)>f\left( x \right) \\

& \Rightarrow f\left( a+h \right)>f\left( a \right) \\

\end{align}\]

So, in this function, if we increase $x$, then $f\left( x \right)$ will increase. So it is an increasing function.

**Note:**Let us take an example of \[f\left( x \right)={{x}^{3}}\]. This is differentiable in nbd of point $a=0$. We remove point $x=0$ from this.

\[f\left( x \right)={{x}^{3}}\]

We differentiate it with respect to $x$.

\[f'\left( x \right)=3{{x}^{2}}\], which is always positive except $x=0$. At $x=0,f'\left( x \right)=0$.

So, $f'\left( 0 \right)=0$, and $f'\left( x \right)>0$ for $x\in R-\left\{ 0 \right\}$.

Hence, \[f\left( x \right)={{x}^{3}}\] is an increasing function.

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