Help: Functions

If $f:\mathbb R\longrightarrow \mathbb R$ satisfying $\left| f\left( x \right) -f\left( y \right) \right| \le { \left| x-y \right| }^{ 3 }$, for all $x,y\in \mathbb R$ and $f\left( 2 \right) =5$, then find $f\left( 4 \right)$.

#Algebra

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

I replaced x by x+h and y by x and took $\lim_{x\rightarrow0}$ i.e $|\lim_{x\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}|\leq0$ and since Modulus of a real no. cannot be less than 0, we get
$f'(x)=0 \Rightarrow f(x)=Constant=C(let)$ And since f(2)=5 $\Rightarrow$ f(x)=5..
Or f(4)=5.
Maybe I have done some mistake as I am lacking time at the moment.. Anyways can you find the error??

Rishabh Jain - 5 years, 4 months ago

Pretty much did the same thing, seems like they misprinted the answer. Thanks for solving.

Akhilesh Prasad - 5 years, 4 months ago

Why must the function be differentiable?

Calvin Lin Staff - 5 years, 3 months ago

@Rishabh Cool,@parv mor. Please if you could post a solution to the question.

Akhilesh Prasad - 5 years, 4 months ago

I am getting f(4)= 5. Is it correct?

Rishabh Jain - 5 years, 4 months ago

That's what i was getting but the answer given is 25. If you could please write your solution so i can check whether my approach is correct.

Akhilesh Prasad - 5 years, 4 months ago

Thanks a lot for solving all my doubts patiently.

Akhilesh Prasad - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$