f(x) = x^2 - 6x + 10 is an increasing functio | vedclass

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(A) To determine the interval where the function $f(x) = x^2 - 6x + 10$ is increasing,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(x^2 - 6x +

Vedclass · Accepted Answer

$(3, \infty)$

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(A) To determine the interval where the function $f(x) = x^2 - 6x + 10$ is increasing,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(x^2 - 6x + 10) = 2x - 6$.$A$ function is increasing when $f'(x) > 0$.$2x - 6 > 0 \implies 2x > 6 \implies x > 3$.Thus,the function is increasing in the interval $(3, \infty)$.Therefore,the correct option is $A$.

$f(x) = x^2 - 6x + 10$ is an increasing function in the . . . . . . interval.

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