The minimum value of the function f(x) = 2x^2 | vedclass

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(B) Given the function $f(x) = 2x^2 - \ln|x|$ for $x \geq 1$.Since $x \geq 1$,we have $|x| = x$,so $f(x) = 2x^2 - \ln x$.Find the derivative: $f'(x) =

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$2$

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(B) Given the function $f(x) = 2x^2 - \ln|x|$ for $x \geq 1$.Since $x \geq 1$,we have $|x| = x$,so $f(x) = 2x^2 - \ln x$.Find the derivative: $f'(x) = 4x - \frac{1}{x}$.For critical points,set $f'(x) = 0$:$4x - \frac{1}{x} = 0 \implies 4x^2 = 1 \implies x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2}$.Since the domain is $x \geq 1$,there are no critical points in the interval $(1, \infty)$.Check the behavior of the function: $f'(x) = \frac{4x^2 - 1}{x}$. For $x > 1$,$4x^2 - 1 > 3$,so $f'(x) > 0$.Since $f'(x) > 0$ for all $x \geq 1$,the function is strictly increasing on $[1, \infty)$.Therefore,the minimum value occurs at the boundary $x = 1$.$f(1) = 2(1)^2 - \ln(1) = 2 - 0 = 2$.

The minimum value of the function $f(x) = 2x^2 - \ln|x|$ for $x \geq 1$ is:

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