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Let $x\in R$,find the maximum of the value $$f(x)=27\sin^8{x}+8\cos^8{x}$$

if let $\sin^2{x}=t$,then $$f=27t^4+8(1-t)^4,t\ge 0$$ without derivative method?

math110
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  • you can already conclude that $f$ is bounded by 35, but this does not answer your question obviously. –  Jan 03 '17 at 15:09

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By setting $t=\sin^2(x)$ we get that the maximum is given by $$ \max_{t\in[0,1]} \left(27t^4+8(1-t)^4\right) $$ but both $t^4$ and $(1-t)^4$ are continuous and convex functions on the interval $[0,1]$, and so it is $g(t)=27t^4+8(1-t)^4$. It follows that $g(t)$ attains its maximum at the boundary of $[0,1]$, namely at $t=1$, and $$\max_{x\in\mathbb{R}} \left(27\sin^8(x)+8\cos^8(x)\right) = \color{red}{27}. $$

Jack D'Aurizio
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  • I think an algerbraic justification of the convex property of these function will make you answer 10 times more satisfactory. – Logan Luther Jan 03 '17 at 15:51
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    @LoganLuther: on the interval $[0,1]$ the function $h(t)=t^4$ is trivially convex, the interested reader may easily found the algebraic justification he needs, if he really needs it. – Jack D'Aurizio Jan 03 '17 at 15:56