Question

Let $f\left(x\right)$ be a real valued function not identically zero such that $f(x+y^n)=f\left(x\right)+{\left\{f\right(y\left)\right\}}^n$ (where $n$ is odd number $>1$) and $f^{\prime }\left(0\right)\ge 0$.

Find out the value of $f^{\prime }\left(10\right)+f\left(5\right)$.

Answer 1

$F(x+(y{)}^n)=F(x)+F(y{)}^n$

put $x=0$ and $y=0$

$F(0+(0{)}^n)=F(0)=F(0)+F(0{)}^n$

$F(0{)}^n=0$

$F\left(0\right)=0$

now put $x=1$ and $y=0$

$F(0+(1{)}^n)=F(1)=F(0)+F(1{)}^n$

$F(1{)}^{(}n-1)=1$

Since $n$ is an odd number & $(n-1)$ is an even number

So, $F\left(1\right)=1$ or $-1$

now put $x=1$ and $y=1$

$F\left(2\right)=F\left(1\right)+F(1{)}^n$

So , $F\left(2\right)=0$ or $2$

In a similar way we can find different different values

put $x=2$ and $y=1$

$F\left(3\right)=1$ or $3$

Similarly,

$F\left(5\right)=3$ or $5$

Now Let's move on to the derivative function

Partially Differentiate the required equation

first with respect to $y$ as a variable

$F(x+y^n{)}^{\prime }=F(x{)}^{\prime }$

Now changing the value of $y>0$ for all real values and keeping $x=0\left(constant\right)$

Then $F(y^n{)}^{\prime }=F(0{)}^{\prime }$

This means that the derivative of the function is constant for all values of $y$

So, from IMVT theorem we have

$F(a{)}^{\prime }=(F\left(3\right)-F\left(2\right)\left)/\right(3-2)$

where $a$ lies somewhere between $2$ and $3$

hence, $F(a{)}^{\prime }=1$

Therefore,

$F(10{)}^{\prime }=1$

Hence Answers are,

$F\left(5\right)=5$ or $3$

$F(10{)}^{\prime }=1$