Let $f: R \rightarrow R$ be a continuous function such that $f(3 x)-f(x)=x$. If $f(8)=7$, then $f(14)$ is equal to:

$\text { Let } f: R \rightarrow R \text { be a continuous function such that } f(3 x)-f(x)=x \text {. If } f(8)=7 \text {, then } f(14) \text { is equal to: }$

 

[Math.LiveDoubtClearing]

Solution: \[ f(3x)-f(x)=x,f(8)=7 \]Put \(x=8,\) \(\Rightarrow f\left(24\right)-f\left(8\right)=8\Rightarrow f\left(24\right)-7=8\)
\[\Rightarrow f\left(24\right)=15\]Let, \(f\left(x\right)=mx+c,\)

Since,
\(f\left(8\right)=7,\) \(7=8m+c\)
\(f\left(24\right)=15,\) \(15=24m+c\)

\(\Rightarrow m=\frac12,c=3\)

So, \[ f\left(x\right)=\frac{x}{2}+3 \]Let's check that \(f\left(x\right)\) is satisfying the given conditions or not.

Given, \[f(3x)-f(x)=x,f(8)=7\]Also,\[f\left(3x\right)-f\left(x\right)=\frac{3x}{2}+3-\frac{x}{2}-3=\frac{3x}{2}-\frac{x}{2}=x,\]\[ f\left(8\right)=\frac82+3=7 \]Hence, \(f\left(x\right)\) is satisfying the given conditions.
Therefore, \[ f\left(x\right)=\frac{x}{2}+3 \]Hence, \[ f\left(14\right)=\frac{14}{2}+3=10 \]

 

[Manish kumar]

Sir why have you assumed f(x) as linear expression?

 

[Math.LiveDoubtClearing]

I assumed to see if \(f(x)\) can be a linear expression or not, and it turned out to be one. If the linear expression did not satisfy the given conditions, then I would try approaching the question in a different way.

 

[Manish kumar]

Sir but why algebraic polynomial why not any other function

 

[Ashutosh]

Sir if it have been turned to not one of the linear expression then what would be our steps to solve it ?

 

[Math.LiveDoubtClearing]

Sir but why algebraic polynomial why not any other function

Since, it is given that the function is continuous and the most easy continuous function that we know is linear function.

 

[Math.LiveDoubtClearing]

Sir if it have been turned to not one of the linear expression then what would be our steps to solve it ?

Then we could try putting other simple functions. It depends on the question.
There can be another method also. Like, in the above question, we you replace \(x\) by $$ \frac{x}{3},\frac{x}{3^2},\ldots,\frac{x}{3^{n}} $$ in the equation $$ f\left(3x\right)-f\left(x\right)=3 $$ You'll get,
$$ f(3x)-f(x)=x $$
$$ f\left(x\right)-f\left(\frac{x}{3}\right)=\frac{x}{3} $$
$$ f\left(\frac{x}{3}\right)-f\left(\frac{x}{3^2}\right)=\frac{x}{3^2} $$
\[\vdots\]
$$ f\left(\frac{x}{3^{n-1}}\right)-f\left(\frac{x}{3^{n}}\right)=\frac{x}{3^{n}} $$

Now add above equations and take the limit as \(n -> \infty\),

$$ \lim_{n\to\infty}\begin{array}{l}f(3x)-f\left(\frac{x}{3^{n}}\right)=x\left(1+\frac{1}{3}+\frac{1}{3^2}+\cdots+\frac{1}{3^{n}}\right)\end{array} $$

$$ f(3x)-f(0)=x\left(\frac{1}{1-\frac13_{}}\right)=\frac{3x}{2} $$

Given, \( f(8) = 7\), Put \(x=8/3\), we get,

\[f(0) = 3\]

\[ \begin{array}{r}f(3 x)=\frac{3 x}{2}+3 \\ f(x)=\frac{x}{2}+3\end{array} \]

 

[Ashutosh]

Thank You sir