In the interval [0, 1],the function f(x) = x^ | vedclass

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(C) Let $f(x) = x^2 - x + 1$. To determine the nature of the function,we find its derivative: $f'(x) = 2x - 1$. We analyze the sign of $f'(x)$ in the

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Neither increasing nor decreasing

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(C) Let $f(x) = x^2 - x + 1$. To determine the nature of the function,we find its derivative: $f'(x) = 2x - 1$. We analyze the sign of $f'(x)$ in the interval $[0, 1]$. At $x = 0$,$f'(0) = 2(0) - 1 = -1 At $x = 1$,$f'(1) = 2(1) - 1 = 1 > 0$. Since the derivative $f'(x)$ changes sign from negative to positive within the interval $[0, 1]$ (specifically,it is zero at $x = 1/2$),the function is neither strictly increasing nor strictly decreasing on the entire interval $[0, 1]$. Therefore,the correct option is $C$.

In the interval $[0, 1]$,the function $f(x) = x^2 - x + 1$ is

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