# If $f\left( x+y \right)=f\left( x \right)+f\left( y \right)\forall x,y\ And f'\left( 0 \right)=2,$ then test the differentiability of $f\left( x \right)$.

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Hint: Since we have to test the differentiability, we have to find the derivative of the function. We can find the derivative of any function using the first principle of derivative and using the functional relation given in the question.

It is given in the question $f\left( x+y \right)=f\left( x \right)+f\left( y \right)\And f'\left( 0 \right)=2,$

In such types of questions, it is necessary to find the function first in order to check its differentiability.

To find the function, we will follow some steps:

(i) Use first principle to find $f'\left( x \right)$

We know by first principle;

\[f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}.............\left( I \right)\]

Since it is given in the question that \[f\left( x+y \right)=f\left( x \right)+f\left( y \right)\], we can write\[f\left( x+h \right)=f\left( x \right)+f\left( h \right)\]. Now, substituting \[f\left( x+h \right)=f\left( x \right)+f\left( h \right)\] in $\left( I \right)$, we get ðŸ¡ª

\[f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)+f\left( h \right)-f\left( x \right)}{h}\]

\[\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)}{h}\]

(ii)Now, we cannot proceed further in step $\left( i \right)$, we proceed to step $\left( ii \right)$ where we will find some boundary value of \[f\left( x \right)\]

$f\left( x+y \right)=f\left( x \right)+f\left( y \right)$

Let us substitute $\text{ }x=y=0$

$\begin{align}

& \Rightarrow f\left( 0+0 \right)=f\left( 0 \right)+f\left( 0 \right) \\

& \Rightarrow f\left( 0 \right)=2f\left( 0 \right) \\

& \Rightarrow f\left( 0 \right)=0 \\

\end{align}$

Now, we will do some changes in step $\left( i \right)$ . From step $\left( ii \right)$, we have;

$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)}{h}$

Adding or subtracting $0$ to any term doesnâ€™t affect the value. Hence, we can write;

$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 0+h \right)-0}{h}$

Since $f\left( 0 \right)=0$, we can replace $0$ by $f\left( 0 \right)$ in the above equation;

$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 0+h \right)-f\left( 0 \right)}{h}.........\left( II \right)$

From the first principle, we know;

$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( h \right)}{h}$

If we substitute $x=0$, we get;

$f'\left( 0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)-f\left( 0 \right)}{h}.........\left( III \right)$

From $\left( III \right)$, substituting $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)-\left( 0 \right)}{h}=f'\left( 0 \right)$in $\left( II \right)$, we get;

$f'\left( x \right)=f'\left( 0 \right)$

It is given in the question $f'\left( 0 \right)=2$

$f'\left( x \right)=2.......\left( IV \right)$

From$\left( IV \right)$, we can find $f'\left( x \right)$ is constant and continuous $\forall x\in R$. This shows that $f\left( x \right)$ is differentiable for $ \ all x\in R$.

To find the function, we will integrate $\left( IV \right)$ with respect to $x$.

\[\begin{align}

& \Rightarrow \int{f'\left( x \right)dx=\int{2dx}} \\

& \Rightarrow f\left( x \right)=2x+C \\

\end{align}\]

Where $C$ is constant of integration.

Note: There is a possibility of mistake while finding the boundary value of the function in step$\left( ii \right)$. The boundary value is to be found by taking help of the information given in the question. For example, since it was given in the question that $f'\left( 0 \right)=2$, that is why we found the value of $f\left( 0 \right)$ in step$\left( ii \right)$.

It is given in the question $f\left( x+y \right)=f\left( x \right)+f\left( y \right)\And f'\left( 0 \right)=2,$

In such types of questions, it is necessary to find the function first in order to check its differentiability.

To find the function, we will follow some steps:

(i) Use first principle to find $f'\left( x \right)$

We know by first principle;

\[f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}.............\left( I \right)\]

Since it is given in the question that \[f\left( x+y \right)=f\left( x \right)+f\left( y \right)\], we can write\[f\left( x+h \right)=f\left( x \right)+f\left( h \right)\]. Now, substituting \[f\left( x+h \right)=f\left( x \right)+f\left( h \right)\] in $\left( I \right)$, we get ðŸ¡ª

\[f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)+f\left( h \right)-f\left( x \right)}{h}\]

\[\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)}{h}\]

(ii)Now, we cannot proceed further in step $\left( i \right)$, we proceed to step $\left( ii \right)$ where we will find some boundary value of \[f\left( x \right)\]

$f\left( x+y \right)=f\left( x \right)+f\left( y \right)$

Let us substitute $\text{ }x=y=0$

$\begin{align}

& \Rightarrow f\left( 0+0 \right)=f\left( 0 \right)+f\left( 0 \right) \\

& \Rightarrow f\left( 0 \right)=2f\left( 0 \right) \\

& \Rightarrow f\left( 0 \right)=0 \\

\end{align}$

Now, we will do some changes in step $\left( i \right)$ . From step $\left( ii \right)$, we have;

$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)}{h}$

Adding or subtracting $0$ to any term doesnâ€™t affect the value. Hence, we can write;

$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 0+h \right)-0}{h}$

Since $f\left( 0 \right)=0$, we can replace $0$ by $f\left( 0 \right)$ in the above equation;

$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 0+h \right)-f\left( 0 \right)}{h}.........\left( II \right)$

From the first principle, we know;

$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( h \right)}{h}$

If we substitute $x=0$, we get;

$f'\left( 0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)-f\left( 0 \right)}{h}.........\left( III \right)$

From $\left( III \right)$, substituting $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)-\left( 0 \right)}{h}=f'\left( 0 \right)$in $\left( II \right)$, we get;

$f'\left( x \right)=f'\left( 0 \right)$

It is given in the question $f'\left( 0 \right)=2$

$f'\left( x \right)=2.......\left( IV \right)$

From$\left( IV \right)$, we can find $f'\left( x \right)$ is constant and continuous $\forall x\in R$. This shows that $f\left( x \right)$ is differentiable for $ \ all x\in R$.

To find the function, we will integrate $\left( IV \right)$ with respect to $x$.

\[\begin{align}

& \Rightarrow \int{f'\left( x \right)dx=\int{2dx}} \\

& \Rightarrow f\left( x \right)=2x+C \\

\end{align}\]

Where $C$ is constant of integration.

Note: There is a possibility of mistake while finding the boundary value of the function in step$\left( ii \right)$. The boundary value is to be found by taking help of the information given in the question. For example, since it was given in the question that $f'\left( 0 \right)=2$, that is why we found the value of $f\left( 0 \right)$ in step$\left( ii \right)$.

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