0
$\begingroup$

Q. If $f:[0,1]\to \mathbb R$ is continues and differentiable. Such that $f(0)= 1$ and $[f(1)]^3 +2f(1) =5$ then prove that there exists a $c$ such that $f'(c)= 2/2+[f'(c)]^2$

How I tried to solve is that $f(0)=1$ and let $f(1)=t$ So $f'(c)= f(1)-f(0)/1-0 =t-1$

Then I tried to eliminate t from the the equation $t^3+2t-5=0$. Initially I was quite confident that it would work but after a bit it doesn't seems to..

pic

$\endgroup$
  • $\begingroup$ What are you trying to solve? $\endgroup$ – k99731 Jan 1 '17 at 8:13
  • $\begingroup$ Sorry I forgot to put the picture $\endgroup$ – Nitish Jan 1 '17 at 8:14
  • $\begingroup$ I think you should try to type it out. $\endgroup$ – k99731 Jan 1 '17 at 8:15
  • $\begingroup$ And now we can see at least one major typo. $\endgroup$ – Did Jan 1 '17 at 11:14
3
$\begingroup$

Consider $$g(x)= \left[f(x) \right]^3 + 2f(x)$$ You can add a constant to this function but it doesn't matter. Then apply the Mean Value Theorem directly to $g(x)$ to obtain \begin{align} g'(c) = \frac{g(1)-g(0)}{1-0} = 2 \end{align} where $c\in[0,1]$. Now look again at the definition of $g(x)$ and use the fact that $g'(c)=2$ to derive an expression for $f'(c)$.

$\endgroup$
1
$\begingroup$

Define $g=2 f(x) +f(x)^3-2x-3$. What is $g(0)$ and $g(1)$?

$\endgroup$
0
$\begingroup$

Let $$g(x)=2f(x)-f^3(x)$$

Apply LMVT , You'll directly get the required result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.