Let f(x)=x^2025-x^2000, x [0,1] and the minimum value of the function f(x) in the interval…

Question

Let $f(x)=x^{2025}-x^{2000}, x \in[0,1]$ and the minimum value of the function $f(x)$ in the interval $[0,1]$ be $(80)^{80}(n)^{-81}$. Then $n$ is equal to

Accepted Answer

$\begin{gathered} f(x)={x}^{2025}-{x}^{2000};x\in [0,1] \ {f}^{'}(x)=2025{x}^{2024}-200{x}^{1999} \ ={x}^{1999}(2025{x}^{5}-2000) \ {f}^{'}(x)=0\Rightarrow {x}^{5}=\frac{80}{81} \ f(x)={({x}^{5})}^{81}-{({x}^{5})}^{80} \ ={(\frac{80}{81})}^{81}-{(\frac{80}{81})}^{80} \ ={(\frac{80}{81})}^{80}(-\frac{1}{81}) \ =(80{)}^{80}(-81{)}^{-81} \end{gathered}$

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