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lucky1940
18.07.2020 •
Mathematics
onsider the equation below. (If an answer does not exist, enter DNE.) f(x) = 8 cos2(x) − 16 sin(x), 0 ≤ x ≤ 2π (a) Find the interval on which f is increasing. (Enter your answer using interval notation.) (No Response) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (No Response) (b) Find the local minimum and maximum values of f. local minimum value (No Response) local maximum value (No Response) (c) Find the inflection points. (x, y) = (No Response) (smaller x-value) (x, y) = (No Response) (larger x-value)
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Ответ:
(a) Increasing:
and Decreasing:![0< x< \frac{\pi}{2}\ \text{or}\ \frac{3\pi}{2}< x< 2\pi](/tpl/images/0709/1147/14601.png)
(b) The local minimum and maximum values are -16 and 16 respectively.
(c) The inflection points are![(\frac{\pi}{6},\ -2)\ \text{and}\ (\frac{5\pi}{6},\ -2)](/tpl/images/0709/1147/3a77f.png)
Step-by-step explanation:
The function provided is:
(a)
Then,![f'(x)=-16cos(x)sin(x)-16cos(x)=-16cos(x)[1+sin(x)]](/tpl/images/0709/1147/8c098.png)
Note,![1+sin(x)\geq 0\ \text{and }\ sin(x)\geq 1\\](/tpl/images/0709/1147/b6cae.png)
Then,
for
.
Also
.
Thus, f (x) is increasing for,
And f (x) is decreasing for,
(b)
From part (a) f (x) changes from decreasing to increasing at
and from increasing to decreasing at
.
The local minimum value is:
The local maximum value is:
(c)
Compute the value of f'' (x) as follows:
So,
And,
Thus, f (x) is concave upward on
and concave downward on
.
If
, then f (x) will be:
If
, then f (x) will be:
The inflection points are
.
Ответ:
ok thank you for that
Step-by-step explanation:
you too have a great day !