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How can I verify that a critical point is a maximum without using the second derivative test?

Here is the specific situation.

There is a function $f(x)\ge 0$ and $x\ge 0$ as they are both distances. Now I found that $$ \frac {df}{dx}=\frac {2}{(1+\frac{2^2}{x^2})x^2}-\frac{8}{(1+\frac{8^2}{x^2}) x^2} $$ and when $\frac {df}{dx}=0, x=\pm 4$, but clearly cannot be $-4$. Now I must show that $x=4$ maximizes $f(x)$ without using the second derivative test.

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  • $\begingroup$ The edit of Subhadeep Dey can´t be right, because $f'(\pm 4) \neq 0$. minusatwelth please improve the question by yourself. $\endgroup$ – callculus Sep 19 '15 at 8:00
  • $\begingroup$ The original post says $df/dx=(2)/((1+(2)^2/x^2) x^2)-(8)/((1+(8)^2/x^2) x^2)$ $\endgroup$ – A.Γ. Sep 19 '15 at 8:00
  • $\begingroup$ Maybe $ df/dx=\dfrac{(2/x^2)}{1+(2/x)^2} - \dfrac{(8/x^2)}{1+(8/x)^2} ? $ OP to clarify.. $\endgroup$ – Narasimham Sep 19 '15 at 8:02
  • $\begingroup$ wolframalpha.com/input/?i=%282%29%2F%28%281%2B%282%29^2%2Fx^2%29+x^2%29-%288%29%2F%28%281%2B%288%29^2%2Fx^2%29+x^2%29%3D0 $\endgroup$ – minusatwelfth Sep 19 '15 at 8:12
  • $\begingroup$ Analyze the derivative before and after that point. $\endgroup$ – Aditya Agarwal Sep 19 '15 at 8:13
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Simplifying the derivative you get $$ f'(x)=\frac{6(16-x^2)}{(x^2+4)(x^2+64)}=\frac{6(4-x)(4+x)}{(x^2+4)(x^2+64)}. $$ Now you see that

  • for $0\le x<4$: $f'(x)>0$ $\Rightarrow$ the function increases,
  • for $x>4$: $f'(x)<0$ $\Rightarrow$ the function decreases.

Hence, $x=4$ is the maximum point.

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  • $\begingroup$ i plotted the graph and it doesn't look increasing on $0<x<4$ see wolframalpha.com/input/?i=%282%29%2F%28%281%2B%282%29^2%2Fx^2%29+x^2%29-%288%29%2F%28%281%2B%288%29^2%2Fx^2%29+x^2%29%3D0 $\endgroup$ – minusatwelfth Sep 19 '15 at 9:17
  • $\begingroup$ @minusatwelfth The link is broken. What is your function? The derivative you gave is the derivative of $\arctan(8/x)-\arctan(2/x)$ and it is increasing for $0\le x\le 4$ (look) $\endgroup$ – A.Γ. Sep 19 '15 at 9:33
  • $\begingroup$ oh ok, how can i show it won't increase again? $\endgroup$ – minusatwelfth Sep 19 '15 at 9:37
  • $\begingroup$ @minusatwelfth After $x=4$ the derivative is negative, so the function decreases. It means that the functions goes down all the way when x goes to infinity so it cannot become larger then $f(4)$, since for that it would need to grow again, but it was not. To look at the derivative sign and calculate the monotonicity intervals is much more efficient way to study max/min than the second derivative (which gives only local extremum). $\endgroup$ – A.Γ. Sep 19 '15 at 9:44

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