Answer
414.6k+ views
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.
$\begin{align}
& f'\left( {{a}^{+}} \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}=finite \\
& f'\left( {{a}^{-}} \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( a-h \right)-f\left( a \right)}{h}=finite \\
\end{align}$ , 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.
![seo images](https://www.vedantu.com/question-sets/ad53dfae-26fb-42d2-98c6-eab1a9d7a71a1196276639826099902.png)
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 –
![seo images](https://www.vedantu.com/question-sets/72260b20-6dbd-4971-ac62-63a16cf430144025425834370723467.png)
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.
![seo images](https://www.vedantu.com/question-sets/75ab2614-f44e-4cdd-bda4-52a7f536250d3268472175291284327.png)
\[f\left( x \right)={{x}^{3}}\]
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)