# Prove the following relation

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..

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

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