The function f(x) = x^3 - 3x^2 + 5x + 7 is | vedclass

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(A) Given the function $f(x) = x^3 - 3x^2 + 5x + 7$. To determine the intervals of increase or decrease,we find the derivative $f'(x)$: $f'(x) = \frac

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increasing in $R$.

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(A) Given the function $f(x) = x^3 - 3x^2 + 5x + 7$. To determine the intervals of increase or decrease,we find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 5x + 7) = 3x^2 - 6x + 5$. Now,we analyze the sign of $f'(x) = 3x^2 - 6x + 5$. We can rewrite this as $f'(x) = 3(x^2 - 2x) + 5 = 3(x^2 - 2x + 1 - 1) + 5 = 3(x-1)^2 - 3 + 5 = 3(x-1)^2 + 2$. Since $(x-1)^2 \ge 0$ for all $x \in R$,it follows that $3(x-1)^2 + 2 \ge 2 > 0$ for all $x \in R$. Since $f'(x) > 0$ for all $x \in R$,the function $f(x)$ is strictly increasing in $R$.

The function $f(x) = x^3 - 3x^2 + 5x + 7$ is

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