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If $f(x)=x^2$ and $g(x)=2\sin x$ then what is the value of $||f-g||_{\infty}=$max$|f(x)-g(x)|$ how can i get value of x where difference of such function has maximum value?

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    @JohnD interval is [0,1]2012-12-17

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The graph of $p(x):=|f(x)-g(x)|$ on $0\le x\le 1$ tells the story:

Mathematica graphics

Computing the maximum numerically, it is approximately $0.8001$ (and occurs at about $x=0.7391$).

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    You want to maximize $p(x)=x^2-2\sin x$ on $[0,1]$. From the graph, you are looking for the interior critical number, i.e. the solution of $p'(x)=2x-2\cos x=0$. Get that with [this](http://www.wolframalpha.com/input/?i=2x-2cos%28x%29%3D0)? Then evaluate $p(x)$ at that $x$ value.2012-12-23