Question

Find the vertical and horizontal asymptotes:

f(x) = x*tan(x)

Answer 1

I know that I could graph it to see the asymptotes, but I have to show the work.

Answer 2

Well, \(\tan(x) \to \pm\infty\) at \(x = \pm \frac{\pi}{2} + 2n\pi\) and so does x*tan(x).
So those are the vertical asymptotes.

Answer 3

For horizontal asymptote, \(\lim x \to \pm \infty ~~x*\tan(x)\) does not exist. In other words it is plus infinity or minus infinity. So no horizontal asymptotes.

Answer 4

The "n" in my first reply is any integer: positive, negative or zero.

Answer 5

Okay, thanks. That makes sense. So, how would I find the intervals of increase or decrease? I know the derivative of my original function is \[x \sec(x)^{2}+\tan(x)\] It's impossible to equal that to zero and solve for it. How would I find the intervals then?

Answer 6

A small typo in my second reply. It should be: In other words it is minus infinity TO plus infinity.
tan(x) is a periodic function and it goes from negative infinity to positive infinity. As x becomes large, x * tan(x) will keep going from minus infinity to plus infinity. The limit does not converge to a specific value. The limit is anywhere from \(-\infty\) to \(+\infty\) and therefore the function does not have a horizontal asymptote.

Answer 7

Intervals of increase / decrease.
\[
f(x) = x\tan(x) \\
f'(x) = x\sec^2(x) + \tan(x) = \frac{x}{\cos^2(x)} + \frac{\sin(x)}{\cos(x)} \\
f'(x) = \frac{x}{\cos^2(x)} + \frac{\sin(x)}{\cos(x)} * \frac{\cos(x)}{\cos(x)}=
\frac{1}{\cos^2(x)}(x+\sin(x)\cos(x)) \\
f'(x) = \sec^2(x) * (x + \frac 12\sin(2x)) \\
\sec^2(x) ~~\text{is never negative due to the squaring. }\\
\text{Therefore, the sign of f '(x) is determined by }(x + \frac 12\sin(2x)) \\
\text{which is positive when x is positive and is negative when x is negative.} \\
\text {Therefore, f(x) increases when x > 0 and decreases when x < 0.}
\]

Answer 8

You can verify that x + 1/2sin(2x) is positive when x is positive and is negative when x is negative as follows. The range of the sine function is -1 to +1. The range of 1/2sin(2x) is -1/2 to +1/2. Initially the sine function is positive when 0 < theta < pi or 0 < 2x < pi which implies 0 < x < pi/2.
Therefore, for 0 < x < pi/2 which is 0 < x < 1.57,
x + 1/2sin(2x) is positive because both x and 1/2*sin(2x) are positive.
As x increases past pi/2 or 1.57, x + 1/2sin(2x) will continue to be positive because the minimum value of 1/2sin(2x) is -1/2 which is not sufficient to make x + 1/2sin(2x) negative for x >= 1.57. So for x > 0, x + 1/2sin(2x) > 0.
Same logic applies for x < 0 where x + 1/2sin(2x) < 0.