Find the product of the min and max value of the function $f(x)=\frac{\sin^4x+\cos^4x}{\sin^6x+\cos^6x}$

Question

Find the product of the min and max value of the function $$f(x)=\dfrac{\sin^4x+\cos^4x}{\sin^6x+\cos^6x}$$ We can simplify the function as follows $$f(x)=\dfrac{(\sin^2x+\cos^2x)-2\sin^2x\cos^2x}{(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)}\\=\dfrac{1-2\sin^2x\cos^2x}{1-3\sin^2x\cos^2x}=\dfrac{1-\frac{(2\sin x\cos x)^2}{4}}{1-\frac{3(2\sin x\cos x)^2}{4}}\\=\dfrac{2-\sin^22x}{2}\cdot\dfrac{4}{4-3\sin^22x}=\dfrac{4-2\sin^22x}{4-3\sin^22x}=\dfrac{4-3\sin^22x+\sin^22x}{4-3\sin^22x}\\=1+\dfrac{\sin^22x}{4-3\sin^22x}$$ How do I continue from here?

Answer 1

Start from your last step:

$$f(x)=1+\dfrac{\sin^22x}{4-3\sin^22x}$$

Case.(1) If $\sin2x=0$, then $f=1$

Case.(2) If $\sin2x\neq 0$, then $0<\sin^22x\le 1$

$$f(x)=1+\dfrac{1}{\frac{4}{\sin^22x}-3}\Rightarrow 1<f\le 2$$

Together with Case.(1), we get:

$$ 1\le f\le 2$$

Answer 2

If you use the multiple angle formula $$f(x)=\frac{\sin^4(x)+\cos^4(x)}{\sin^6(x)+\cos^6(x)}=\frac{2 (\cos (4 x)+3)}{3 \cos (4 x)+5}$$ Therefore using $\quad -1\leq \cos(x) \leq 1$ $$1 \leq f(x) \leq 2$$

Answer 3

Now, for each $x\in[0,1]$, let $g(x)=1+\dfrac x{4-3x}$. Then, for each $x\in[0,1]$,$$g'(x)=\frac4{(4-3x)^2}>0,$$and therefore $g$ is strictly increasing. So, if $x\in[0,1]$, $g(0)\leqslant g(x)\leqslant g(1)$; in other words, $1\leqslant g(x)\leqslant2$. So, since $f(x)=g\left(\sin^2(2x)\right)$, you always have $1\leqslant f(x)\leqslant2$. But $f(0)=1$ and $f(1)=2$. It follows that $\min f=1$ and that $\max f=2$.

Answer 4

As an alternative, for $\cos x \neq 0$ with $t=\tan^2 x \ge 0$ we have

$$f(x)=\dfrac{\sin^4x+\cos^4x}{\sin^6x+\cos^6x} (\sin^2x+\cos^2x)=1+\dfrac{\sin^4x\cos^2 x+\cos^4x\sin^2 x}{\sin^6x+\cos^6x}=$$

$$= 1+\dfrac{\tan^4 x+\tan^2 x}{\tan^6x+1} \implies f(t)=1+\frac{t^2+t}{t^3+1}=1+\frac1{t+\frac1t -1}$$

and then since $t+\frac1t \in [2,\infty)$ with minimum value for $t=\frac1t =1$ we have

$$1\le f(x)\le2$$