The function $f(x)$ is continuous and differentiable in $[0,1]$ if $f'(x)\le 10$ for all $x\in[0,1]$ and $f(0)=0$,

What is the maximum possible value of $f(x)$ for $x\in [0,1]$ ?

Any help would be greatly appreciated, thanks.


closed as off-topic by Henrik, Christopher, Gibbs, José Carlos Santos, Namaste Nov 12 '18 at 14:48

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    $\begingroup$ Just apply Mean Value Theorem. $\endgroup$ – Kavi Rama Murthy Nov 12 '18 at 10:08
  • $\begingroup$ Welcome to math.SE. Questions like yours that don't include the authors own work/thoughts are unpopular here, and tend to get closed quite fast. You should edit the question to include that. When editing the question you should also use some MathJax (LaTeX) for formatting, the help center has links to get you started with that. $\endgroup$ – Henrik Nov 12 '18 at 10:32

$\forall x\in [0,1]$ we have,

$\frac{f(x)-f(0)}{x-0}\le 10$ {By lagrange's mean value theorem we have a $c\in [0,x]\ such\ that\ \frac{f(x)-f(0)}{x-0}=f'(c)\le10$}

$\therefore$ maximum value of f(x) is 10$x$ in [0,1].

Hope it helps:)


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