The range of $f(x)=\cos^{2n}(x)+\sin^{2n}(x)$

Question

I have $f(x)=\cos^{2n}(x)+\sin^{2n}(x),\; n\in \mathbb{N},\; n\geq 2,\;x\in \mathbb{R}$ I need to find the range of the function. I took $n=2$ and I got $f(x)=1-\frac{1}{2}\sin^{2}(2x)$ and the range of this is $[1/2,1]$

Also, for $n=3$ I got $[1/4,1]$.

How to find the range for $n$?

Answer 1

With $t=\cos^2(x)$ such that $0\le t\le 1$, you bracket

$$t^n+(1-t)^n.$$

The stationary points are the roots of

$$t^{n-1}-(1-t)^{n-1}=0$$ or $$t=\frac12.$$

Hence,

$$f(x)\in[2^{1-n},1].$$

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Without derivatives:

$$t^n+(1-t)^n$$ is obviously symmetric around $t=\dfrac12$, and with $s:=t+\dfrac12$,

$$\left(\frac12+s\right)^n+\left(\frac12-s\right)^n.$$

If we develop using the binomial formula, only even degree terms will remain so that the polynomial is monotonic for $s>0$. Hence the minimum is achieved for $s=0$, and the maximum for $s=\dfrac12$.

Answer 2

Hint:

You have to find the range of the function $f(t)=t^n+(1-t)^n,\quad t\in[0,1]$