The function f(x) = 2x^3 - 3x^2 + 90x + 174 i | vedclass

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(D) To determine the interval where the function $f(x) = 2x^3 - 3x^2 + 90x + 174$ is increasing,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2

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$-\infty < x < \infty$

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(D) To determine the interval where the function $f(x) = 2x^3 - 3x^2 + 90x + 174$ is increasing,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2x^3 - 3x^2 + 90x + 174) = 6x^2 - 6x + 90$.We can factor out $6$ from the expression: $f'(x) = 6(x^2 - x + 15)$.For the function to be increasing,we require $f'(x) > 0$,which means $6(x^2 - x + 15) > 0$,or $x^2 - x + 15 > 0$.We examine the discriminant $D$ of the quadratic $x^2 - x + 15$. Here $a = 1, b = -1, c = 15$.$D = b^2 - 4ac = (-1)^2 - 4(1)(15) = 1 - 60 = -59$.Since the discriminant $D  0)$,the quadratic $x^2 - x + 15$ is always positive for all real values of $x$.Therefore,$f'(x) > 0$ for all $x \in (-\infty, \infty)$.Thus,the function is increasing on the interval $(-\infty, \infty)$.

The function $f(x) = 2x^3 - 3x^2 + 90x + 174$ is increasing in the interval

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