solve |2x - 6| > 10 a. {x|x < -8 or x > 2} b. {x|x < -2 or x > 8} c. {x|-2 < x < 8}

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Consider the following function. f(x) = 2x3 + 9x2 − 24x (a) Find the critical numbers of f. (Enter your answers as a comma-separ

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Answer:

(a) The critical number of $f(x)$ are $x=-4, 1$

(b)

Increasing for $(-\infty, -4)$
Decreasing for $(-4, 1)$
Increasing for $(1, \infty)$

(c)

relative maximum $(-4, 112)$
relative minimum $(1, -13)$

Step-by-step explanation:

(a) The critical numbers of a function are given by finding the roots of the first derivative of the function or the values where the first derivative does not exist. Since the function is a polynomial, its domain and the domain of its derivatives is $(-\infty, \infty)$ . Thus:

$\frac{df(x)}{dx} = \frac{d(2x^3+9x^2-24x)}{dx} =6 x^2+18x -24\\6 x^2+18x -24=0\\\boxed{x=-4, x=1}$

(b)

A function $f(x)$ defined on an interval is monotone increasing on $(a, b)$ if for every $x_1, x_2 \in (a, b): x_1$ implies $f(x_1)$

A function $f(x)$ defined on an interval is monotone decreasing on $(a, b)$ if for every $x_1, x_2 \in (a, b): x_1$ implies $f(x_1)>f(x_2)$

Combining the domain $(-\infty, \infty)$ with the critical numbers we have the intervals $(-\infty, -4)$ , $(-4, 1)$ and $(1, \infty)$ . Note that any of the points are included, in the case of the infinity it is by definition and the critical number are never included because the function monotony is not defined in the critical points, i.e. it is not monotone increasing or decreasing. Now, let's check for the monotony in each interval, for this, we check for the sign of the first derivative in each interval. Evaluating in each interval the first derivative (one point is enough), we obtain the monotony of the function to be:

Increasing for $(-\infty, -4)$
Decreasing for $(-4, 1)$
Increasing for $(1, \infty)$

(c) From the values obtained in (a) so the relative extremum are the points $(-4, 112)$ and $(1, -13)$ . The $y$ -values are found by evaluating the critical numbers in the original function. Since the first derivative decreases after passing through $x=-4$ and increases after passing through the point $x=1$ we have: