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I was asked to find the minimum and maximum values ​​of the functions:

  1. $y=\sin^2x/(1+\cos^2x)$;
  2. $y=\sin^2x-\cos^4x$.

What I did so far:

  1. $y' = 2\sin(2x)/(1+\cos^2x)^2$
    How do I check if they are suspicious extrema points? After this function is cyclical and therefore only section that is not $(-\infty,\infty)$ can there be a local minimum/maximum.

  2. $y' = \sin(2x)+4\cos^3(x)\cdot\sin(x)$

Any suggestions?

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2 Answers 2

up vote 1 down vote accepted

Hint: $y=f(x)$ has maximum or minimum when $y'=0$.

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I know but the derivative is equal to 0 every cycle .. –  Ofir Attia Apr 17 '13 at 12:54
    
That means there are maximum ir minimum value for each cycle. –  ᴊ ᴀ s ᴏ ɴ Apr 17 '13 at 12:57
    
desmos.com/calculator/ztqwffwrp9 @ desmos.com –  Ofir Attia Apr 17 '13 at 13:04
    
both of them are equal to zero every pi/2 right? –  Ofir Attia Apr 17 '13 at 13:05
    
Yeah, you're right. –  ᴊ ᴀ s ᴏ ɴ Apr 17 '13 at 13:10

If you don't really need to use derivatives,

$1.$ Clearly, $y=\frac{\sin^2x}{1+\cos^2x}\ge 0$ in fact $=0$ if $\sin x=0$

$y-1=\frac{\sin^2x}{1+\cos^2x}-1=\frac{\sin^2x-(1+\cos^2x)}{1+\cos^2x}=-2\frac{\cos^2x}{1+\cos^2x}\le 0$ in fact $=0$ if $\cos x=0$

$\implies y-1\le 0\iff y\le 1\implies 0\le y\le 1$

$2.$ $$y=\sin^2x-\cos^4x=1-\cos^2x-\cos^4x=1-\cos^2x(1+\cos^2x)$$

$$\text{Now, }0\le \cos^2x\le1\implies 1\le\cos^2x+1\le2 \implies 0\le \cos^2x(\cos^2x+1)\le2$$

$$\implies 0\ge-\cos^2x(\cos^2x+1)\ge-2\iff 1\ge1-\cos^2x(\cos^2x+1)\ge-1$$

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