The range of values of x for which f(x)=x^3+6 | vedclass

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(A) Given function: $f(x) = x^3 + 6x^2 - 36x + 7$ Find the derivative: $f'(x) = 3x^2 + 12x - 36$ Factor the derivative: $f'(x) = 3(x^2 + 4x - 12) = 3(

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$(-\infty, -6) \cup (2, \infty)$

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(A) Given function: $f(x) = x^3 + 6x^2 - 36x + 7$ Find the derivative: $f'(x) = 3x^2 + 12x - 36$ Factor the derivative: $f'(x) = 3(x^2 + 4x - 12) = 3(x + 6)(x - 2)$ For the function to be increasing,we require $f'(x) > 0$: $3(x + 6)(x - 2) > 0$ $(x + 6)(x - 2) > 0$ Using the sign scheme for the quadratic inequality,the expression is positive when $x  2$. Thus,the range of values is $x \in (-\infty, -6) \cup (2, \infty)$.

The range of values of $x$ for which $f(x)=x^3+6x^2-36x+7$ is increasing is:

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