For all values of x,the function f(x) = 2x^3 | vedclass

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(A) To determine the nature of the function,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2x^3 + 6x^2 + 7x - 19) = 6x^2 + 12x + 7$.We can rewri

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Increasing

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(A) To determine the nature of the function,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2x^3 + 6x^2 + 7x - 19) = 6x^2 + 12x + 7$.We can rewrite this as $f'(x) = 6(x^2 + 2x) + 7$.Completing the square,$f'(x) = 6(x^2 + 2x + 1 - 1) + 7 = 6(x + 1)^2 - 6 + 7 = 6(x + 1)^2 + 1$.Since $(x + 1)^2 \geq 0$ for all real $x$,it follows that $6(x + 1)^2 + 1 \geq 1$.Thus,$f'(x) > 0$ for all $x \in \mathbb{R}$.Since the derivative is strictly positive for all $x$,the function $f(x)$ is strictly increasing.

For all values of $x$,the function $f(x) = 2x^3 + 6x^2 + 7x - 19$ is:

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