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Step 3 We have determined that in the interval $0 \le x \le 2\pi$, $f'(x) = 0$ when $x = \frac{\pi}{4}$ and when $x = \frac{5\pi}{4}$. This allows us to construct a table to investigate the value of $f'(x)$ over our intervals of interest. To complete the third column, refer again to the increasing/decreasing test with respect to a function $f$. • If $f'(x) > 0$ on an interval, then $f$ is increasing on that interval. • If $f'(x) < 0$ on an interval, then $f$ is decreasing on that interval. $x$ $f'(x) = 6\cos(x) - 6\sin(x)$ $f(x)$ $(0, \frac{\pi}{4})$ --Select--- --Select--- $(\frac{\pi}{4}, \frac{5\pi}{4})$ --Select--- --Select--- $(\frac{5\pi}{4}, 2\pi)$ --Select--- --Select--- Using the table above, we can now state the interval(s) in $0 \le x \le 2\pi$ on which $f$ is increasing. (Enter your answer using interval notation.)

          Step 3
We have determined that in the interval $0 \le x \le 2\pi$, $f'(x) = 0$ when $x = \frac{\pi}{4}$ and when $x = \frac{5\pi}{4}$. This allows us to construct a table to investigate the value of $f'(x)$ over our intervals of interest. To complete the third column, refer again to the increasing/decreasing test with respect to a function $f$.
• If $f'(x) > 0$ on an interval, then $f$ is increasing on that interval.
• If $f'(x) < 0$ on an interval, then $f$ is decreasing on that interval.
$x$	$f'(x) = 6\cos(x) - 6\sin(x)$	$f(x)$
$(0, \frac{\pi}{4})$	--Select---	--Select---
$(\frac{\pi}{4}, \frac{5\pi}{4})$	--Select---	--Select---
$(\frac{5\pi}{4}, 2\pi)$	--Select---	--Select---
Using the table above, we can now state the interval(s) in $0 \le x \le 2\pi$ on which $f$ is increasing. (Enter your answer using interval notation.)

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Step 3
We have determined that in the interval 0 ≤ x ≤ 2π, f'(x) = 0 when x = (π)/(4) and when x = (5π)/(4). This allows us to construct a table to investigate the value of f'(x) over our intervals of interest. To complete the third column, refer again to the increasing/decreasing test with respect to a function f.
• If f'(x) > 0 on an interval, then f is increasing on that interval.
• If f'(x) < 0 on an interval, then f is decreasing on that interval.
x f'(x) = 6cos(x) - 6sin(x) f(x)
(0, (π)/(4)) –Select— –Select—
((π)/(4), (5π)/(4)) –Select— –Select—
((5π)/(4), 2π) –Select— –Select—
Using the table above, we can now state the interval(s) in 0 ≤ x ≤ 2π on which f is increasing. (Enter your answer using interval notation.)

Added by Anna H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Suzanne W.

08/14/2022

Step 1

We are given that $f'(x) = 0$ when $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. We need to determine the sign of $f'(x)$ in the intervals $(0, \frac{\pi}{4})$, $(\frac{\pi}{4}, \frac{5\pi}{4})$, and $(\frac{5\pi}{4}, 2\pi)$. Show more…

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Step 3 We have determined that in the interval $0 \le x \le 2\pi$, $f'(x) = 0$ when $x = \frac{\pi}{4}$ and when $x = \frac{5\pi}{4}$. This allows us to construct a table to investigate the value of $f'(x)$ over our intervals of interest. To complete the third column, refer again to the increasing/decreasing test with respect to a function $f$. • If $f'(x) > 0$ on an interval, then $f$ is increasing on that interval. • If $f'(x) < 0$ on an interval, then $f$ is decreasing on that interval. x $f'(x) = 6\cos(x) - 6\sin(x)$ $f(x)$ $(0, \frac{\pi}{4})$ --Select-- --Select-- $(\frac{\pi}{4}, \frac{5\pi}{4})$ --Select-- --Select-- $(\frac{5\pi}{4}, 2\pi)$ --Select-- --Select-- Using the table above, we can now state the interval(s) in $0 \le x \le 2\pi$ on which $f$ is increasing. (Enter your answer using interval notation.)

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00:02 Okay, so you have the function 6 cosine squared of x minus 12 sine of x.

00:16 So first we are going to take the first derivative.

00:22 The first derivative 6 times 2 is 12, then cosine of x, and then the derivative of cosine is negative sine, then minus 12 cosine of x.

00:45 So we can simplify this.

00:47 This is negative 12 cosine of x times sine of x plus one.

01:01 This is equal to zero.

01:04 So to find the critical points, this is equal to zero when cosine is equal to zero, sign is equal to negative one.

01:14 So that is pi over two and three pi over two.

01:21 So in the interval, pi over 2 and 3 pi over 2, so we can find increasing and decreasing.

01:38 From 0 to pi over 2, the derivative is negative.

01:45 From pi over 2 to 3 pi over 2, the derivative is positive.

01:49 And from 3 pi over 2 to 2 pi, the derivative is negative.

01:53 So it is increasing from, sorry, it is increasing from pi over two to three pi over two.

02:06 It is decreasing from zero to pi over two and then from three pi over two to two pi.

02:20 We have those two values to find maximums and minimums.

02:24 Since we're going from decreasing to increasing pi over 2, i'll write it above it, is a minimum, and 3 pi over 2 is a maximum.

02:39 And then we need to find the inflection points.

02:42 To find the inflection points, we need to take the second derivative...

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