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?
What is maximum value of function?
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calculus
real-analysis
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4Try taking the derivative and finding where the max/mins occur. You can determine it from this point (but if the interval is unbounded, it should be clear that it is $\infty$. – 2012-12-16
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2Max over what interval? – 2012-12-16
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0$g$ is bounded but $f$ is not. What does that tell you? – 2012-12-16
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0Taking the derivative is not really appropriate here (apart from minor details like differentiability which can be handled by squaring); you need to establish that a maximum exists first. Setting the derivative to zero will, at best, locate a local maximum in $[0,\frac{1}{2}]$. – 2012-12-16
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1As explicitely asked by @JohnD: **over which interval**? For the most natural choice, there is no maximum. – 2012-12-16
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0@JohnD interval is [0,1] – 2012-12-17
1 Answers
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The graph of $p(x):=|f(x)-g(x)|$ on $0\le x\le 1$ tells the story:

Computing the maximum numerically, it is approximately $0.8001$ (and occurs at about $x=0.7391$).
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0how can i calculate? – 2012-12-23
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0You 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