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3 years ago

Is my answer correct? Please see attachment and notify me.

Mathematics

2 answers:

statuscvo [17]3 years ago

6 0

Last time I checked that was about right, but it's been a few years since then.

Oksana_A [137]3 years ago

3 0

Your answer looks correct.

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Answer the following question about the function whose derivative is given below

Komok [63]

Answer:

a) The critical points are $x = 3$ and $x = -6$ .

b) f is decreasing in the interval $(-\infty, -6)$

f is increasing in the intervals $(-6,3)$ and $(3,\infty)$ .

c) Local minima: $x = -6$

Local maxima: No local maxima

Step-by-step explanation:

(a) what are the critical points of f?

The critical points of f are those in which $f^{\prime}(x) = 0$ . So

$f^{\prime}(x) = 0$

$(x-3)^{2}(x+6) = 0$

So, the critical points are $x = 3$ and $x = -6$ .

(b) on what intervals is f increasing or decreasing? (if there is no interval put no interval)

For any interval, if $f^{\prime}$ is positive, f is increasing in the interval. If it is negative, f is decreasing in the interval.

Our critical points are $x = 3$ and $x = -6$ . So we have those following intervals:

$(-\infty, -6), (-6,3), (3, \infty)$

We select a point x in each interval, and calculate $f^{\prime}(x)$ .

-------------------------

$(-\infty, -6)$

$f^{\prime}(-7) = (-7-3)^{2}(-7+6) = (100)(-1) = -100$

f is decreasing in the interval $(-\infty, -6)$

---------------------------

$(-6,3)$

$f^{\prime}(2) = (2-3)^{2}(2+6) = (1)(8) = 8$

f is increasing in the interval $(-6,3)$ .

------------------------------

$(3, \infty)$

$f^{\prime}(4) = (4-3)^{2}(4+6) = (1)(10) = 10$

f is increasing in the interval $(3,\infty)$ .

(c) At what points, if any, does f assume local maximum and minima values. ( if there is no local maxima put mo local maxima) if there is no local minima put no local minima

At a critical point x, if the function goes from decreasing to increasing, it is a local minima. And if the function goes from increasing to decreasing, it is a local maxima.

So, for each critical point is this problem:

At $x = -6$ , f goes from decreasing to increasing.

So $x = -6$ , f assume a local minima value

At $x = 3$ , f goes from increasing to increasing. So, there it is not a local maxima nor a local minima. So, there is no local maxima for this function.

4 0

3 years ago

What is the equation of the line that is

Fantom [35]

Answer:

$y = m_{AB}(X-2)-1$

Step-by-step explanation:

When two lines are parallel means that their slopes are equal. Therefore the line AB will have same slope to a parallel second line, $m_{AB} = m_{2}$ .