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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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, Christopher, Gibbs, José Carlos Santos, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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
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$\forall x\in [0,1]$ 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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