f(x) = x^3 - 27x + 5 is an increasing functio | vedclass

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(B) function $f(x)$ is increasing if $f'(x) > 0$.Given $f(x) = x^3 - 27x + 5$.Differentiating with respect to $x$,we get $f'(x) = 3x^2 - 27$.For the f

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$|x| > 3$

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(B) function $f(x)$ is increasing if $f'(x) > 0$.Given $f(x) = x^3 - 27x + 5$.Differentiating with respect to $x$,we get $f'(x) = 3x^2 - 27$.For the function to be increasing,we set $f'(x) > 0$:$3x^2 - 27 > 0$$3(x^2 - 9) > 0$$x^2 - 9 > 0$$x^2 > 9$Taking the square root on both sides,we get $|x| > 3$.Thus,the function is increasing when $|x| > 3$.

$f(x) = x^3 - 27x + 5$ is an increasing function,when

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