The function f(x) = x^3 - 3x^2 - 24x + 5 is a | vedclass

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(A) Given function is $f(x) = x^3 - 3x^2 - 24x + 5$.To find the intervals where the function is increasing,we calculate the derivative $f'(x)$:$f'(x)

Vedclass · Accepted Answer

$( - \infty, -2) \cup (4, \infty)$

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(A) Given function is $f(x) = x^3 - 3x^2 - 24x + 5$.To find the intervals where the function is increasing,we calculate the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x^3 - 3x^2 - 24x + 5) = 3x^2 - 6x - 24$.For the function to be increasing,we must have $f'(x) > 0$:$3x^2 - 6x - 24 > 0$.Dividing the inequality by $3$,we get:$x^2 - 2x - 8 > 0$.Factoring the quadratic expression:$x^2 - 4x + 2x - 8 > 0$$(x - 4)(x + 2) > 0$.Using the sign scheme method,the product $(x - 4)(x + 2)$ is positive when $x  4$.Therefore,the function is increasing in the interval $( - \infty, -2) \cup (4, \infty)$.

The function $f(x) = x^3 - 3x^2 - 24x + 5$ is an increasing function in the interval given below:

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