Find the absolute maximum of x^{40}-x^{20} on | vedclass

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(B) Let $f(x) = x^{40} - x^{20}$ on the interval $[0, 1]$.To find the absolute maximum,we first find the critical points by setting the derivative $f'

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Let $f(x) = x^{40} - x^{20}$ on the interval $[0, 1]$.To find the absolute maximum,we first find the critical points by setting the derivative $f'(x) = 0$.$f'(x) = 40x^{39} - 20x^{19} = 20x^{19}(2x^{20} - 1)$.Setting $f'(x) = 0$,we get $x = 0$ or $2x^{20} = 1$,which implies $x^{20} = \frac{1}{2}$,so $x = (\frac{1}{2})^{1/20}$.Now,evaluate $f(x)$ at the critical point and the endpoints $x=0$ and $x=1$:$f(0) = 0^{40} - 0^{20} = 0$.$f(1) = 1^{40} - 1^{20} = 1 - 1 = 0$.$f((\frac{1}{2})^{1/20}) = ((\frac{1}{2})^{1/20})^{40} - ((\frac{1}{2})^{1/20})^{20} = (\frac{1}{2})^2 - (\frac{1}{2})^1 = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$.Comparing the values $f(0)=0$,$f(1)=0$,and $f((\frac{1}{2})^{1/20}) = -\frac{1}{4}$,the absolute maximum value is $0$.

Find the absolute maximum of $x^{40}-x^{20}$ on the interval $[0,1]$.

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