Differentiation of Function and Relation - Practice Questions & MCQ

Functional Equations

To solve the question of the type where functional equation in two independent variables with some conditions are given it is asked to find the derivative of the function at some value of x or it is asked to find the function. 

Let’s go through some illustrations to understand how to deal with such questions.

Illustration 1

Let $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ for all real $x$ and $y$. If $f^{\prime}(0)$ exists and equal to -1 and $f(0)=1$. Then $f^{\prime}(x)$ is

Solution
Given equation is $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$
Putting $y=0$ and $f(0)=1$ in (i), we have,

$
f\left(\frac{x}{2}\right)=\frac{1}{2}[f(x)+1] \quad \Rightarrow f(x)=2 f\left(\frac{x}{2}\right)-1
$
Now,

$
\begin{aligned}
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
&=\lim _{h \rightarrow 0} \frac{f\left(\frac{2 x+2 h}{2}\right)-f(x)}{h} \\
&=\lim _{h \rightarrow 0} \frac{\frac{f(2 x)+f(2 h)}{2}-f(x)}{h} \\
&=\lim _{h \rightarrow 0} \frac{f(2 x)+f(2 h)-2 f(x)}{2 h} \\
&=\lim _{h \rightarrow 0} \frac{2 f(x)-1+f(2 h)-2 f(x)}{2 h} \quad \text { [ using (i)] } \\
&=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(2 \mathrm{~h})-1}{2 \mathrm{~h}}=\mathrm{f}^{\prime}(0) \\
& \therefore \mathrm{f}^{\prime}(\mathrm{x})=-1
\end{aligned}
$
If $f(x)$ is asked then we can integrate this equation obtained.