For the function f(x) = x^{40} - x^{20},find | vedclass

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

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$-1/4$

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(B) To find the absolute minimum of $f(x) = x^{40} - x^{20}$ on the interval $[0, 1]$,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 $x^{20} = 1/2$,so $x = (1/2)^{1/20}$.Since $(1/2)^{1/20}$ is in the interval $(0, 1)$,we evaluate $f(x)$ at the critical point and the endpoints $x = 0$ and $x = 1$.$f(0) = 0^{40} - 0^{20} = 0$.$f(1) = 1^{40} - 1^{20} = 0$.$f((1/2)^{1/20}) = ((1/2)^{1/20})^{40} - ((1/2)^{1/20})^{20} = (1/2)^2 - (1/2)^1 = 1/4 - 1/2 = -1/4$.Comparing the values $0, 0,$ and $-1/4$,the absolute minimum value is $-1/4$.

For the function $f(x) = x^{40} - x^{20}$,find the absolute minimum value in the interval $[0, 1]$.

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