The real valued function f(x) = frac{x^2}{2} | vedclass

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(B) To determine the intervals of increase and decrease, we find the derivative $f'(x)$.$f'(x) = \frac{d}{dx} \left( \frac{x^2}{2} - \log(x^2 + x + 1)

Vedclass · Accepted Answer

Strictly increasing in $(1, \infty)$

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(B) To determine the intervals of increase and decrease, we find the derivative $f'(x)$.$f'(x) = \frac{d}{dx} \left( \frac{x^2}{2} - \log(x^2 + x + 1) \right) = x - \frac{2x + 1}{x^2 + x + 1}$.$f'(x) = \frac{x(x^2 + x + 1) - (2x + 1)}{x^2 + x + 1} = \frac{x^3 + x^2 + x - 2x - 1}{x^2 + x + 1} = \frac{x^3 + x^2 - x - 1}{x^2 + x + 1}$.Factoring the numerator: $x^2(x + 1) - 1(x + 1) = (x^2 - 1)(x + 1) = (x - 1)(x + 1)^2$.So, $f'(x) = \frac{(x - 1)(x + 1)^2}{x^2 + x + 1}$.Since $x^2 + x + 1 > 0$ and $(x + 1)^2 \ge 0$ for all real $x$, the sign of $f'(x)$ depends on $(x - 1)$.$f'(x) > 0$ when $x > 1$, so the function is strictly increasing in $(1, \infty)$.Thus, option $B$ is correct.

The real valued function $f(x) = \frac{x^2}{2} - \log(x^2 + x + 1)$ is

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