Question

If the function $f\left(x\right)$ is defined by $f\left(x\right)=a+bx$ and $f^r=fff....$ (repeated $r$ times), then $\frac{d}{dx}\left\{f^r\right(x\left)\right\}$ is equal to

• A. $a+b^{r}x$
• B. $ar+b^{r}x$
• C. $ar$
• D. $b^r$

Answer 1

The correct option is D $b^r$

Given, $f\left(x\right)=a+bx$
$f\left\{f\right(x\left)\right\}=a+b(a+bx)$
$=ab+a+b^{2}x=a(1+b)+b^{2}x$
$f\left[f\left\{f\right(x\left)\right\}\right]=f\left\{a\right(1+b)+b^{2}x\}$
$=a+b\left\{a\right(1+b)+b^{2}x\}$
$=a(1+b+b^2)+b^{3}x$
Therefore, $f^r\left(x\right)=a(1+b+b^2+....+b^{r-1})+b^{r}x$
$=a\left(\frac{b^r-1}{b-1}\right]+b^{r}x$
$\Rightarrow \frac{d}{dx}\left\{f^{r}x\right\}=b^r$