Find the absolute minimum and absolute maximum in [-1,2] of the function $ \ f(x)=|x| \ $. $$ $$ I have got absolute minimum at x=0 , ie., $ f(0)=0 $ is the absolute mimimum. The absolute maximum is $ f(2)=2 $ . Is this correct ?
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2Yes. $\phantom{}$ – Umberto P. May 23 '17 at 16:05
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Isn't the answer quite obvious (I mean it comes intuitively)... – Soham May 23 '17 at 16:12
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We can't differentiate normally when there is a modulus sign, so consider $g(x)=f(x)^2=|x|^2=x^2$
Then $g'(x) =2x$
Setting $g'(x)=0\implies f(x)^2=0\implies f(x)=0$ so clearly a minimum there.
Then we can maximise $g(x)$ by testing the upper and lower bounds of the given domain i.e. $g(-1)$ and $g(2)$.
$g(-1)=1<g(2) =4$. Thus $g(x)$ is maximised as $x=2$, which implies that $f(x)$ is maximised at $x=2$, so $f_{\text{max}}=f(2)=2$
Alternatively you could consider the separate cases when $f(x)$ is positive or negative and compute the derivatives accordingly, however the method I have used is often more helpful when dealing with more difficult questions.
mrnovice
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