Question

Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x(a) Find the intervals on which f is increasing. (Enter the interval that contains smaller numbers first.)( , ) ∪ ( , )Find the interval on which f is decreasing.( , )(b) Find the local minimum and maximum values of f.(min)-(max)-(c) Find the inflection point.( , )Find the interval on which f is concave up.( , )Find the interval on which f is concave down.( , )

Answer 1

Answer:

(a) The function is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is $x = -\frac{1}{2}$

(d) The function is concave upward on $\left(- \frac{1}{2}, \infty\right)$ and concave downward on $\left(-\infty, - \frac{1}{2}\right)$

Step-by-step explanation:

(a) To find the intervals where $f(x) = 2x^3 + 3x^2 -180x$ is increasing or decreasing you must:

1. Differentiate the function

$\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180$

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

$f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6$

These points divide the number line into three intervals:

$(-\infty,-6)$, $(-6,5)$, and $(5, \infty)$

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

$\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right$

Therefore f(x) is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

• f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.

• f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

$f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6$

We set f''(x) = 0

$f''(x) =12x+6 =0\\\\x=-\frac{1}{2}$

Analyzing concavity, we get

$\left\begin{array}{cccc}Interval&x-value&f''(x)\\(-\infty,-1/2)&-2&-18\\(-1/2,\infty)&0&6\\\end{array}\right$

The function is concave upward on $(-1/2,\infty)$ because the f''(x) > 0 and concave downward on $(-\infty,-1/2)$ because the f''(x) < 0.

f(x) is concave down before $x = -\frac{1}{2}$, concave up after it. So f(x) has an inflection point at $x = -\frac{1}{2}$.