# Test review: intervals of increasing/decrease and L'Hôpital's rule

I am not sure what I am doing wrong, but I just got this whole test wrong except one question.

Anyways I was having trouble with

1) Find the intervals on which $f(x)=x-2\cos x$, $0\leq x \leq 2\pi$,

• a) is increasing;
• b) decreasing.

I got $(\pi/6,\infty)$ increasing and $(-\infty, 0)$ decreasing. This is of course wrong but I don't quite understand what I was supposed to do differently. I forgot radians and had to figure it all out by hand because I don't have the unit circle memorized, this probably took me about half the test time (maybe an hour).

3) For the function $f(x) = (x+1)/x$, where is the graph

• a) concave up
• b) concave down

I got $0$ and $4$. I know this is wrong but I don't really know where I went wrong. I found the derivative and critical numbers $(2-x)/x^2$ and then I found the critical numbers, $-2$ then I found the derivative of that and got a critical number of $0$ and $4$ and found it was concave up on zero to $4$ and that is it.

7) Find the limit of $(\sec x-\tan x)$ as $x$ approached $\pi/2$.

I worked out $1/\cos x - \sin x/\cos x$ then $(1-\sin x)/\cos x$ and then $-\cos x/-\sin x = 0$.

I could pretty much post the entire test on here but I don't want to waste anyone's time.

For (1), you don't tell us what you did, so it is hard to say where you may have gone wrong.

Here's how to go about it: we have $f(x) = x- 2\cos x$ on $[0,2\pi]$. First, we determine the points where the derivative could change sign (the critical points):

Since $f'(x) = 1 + 2 \sin (x)$, the critical points are the points where $f'(x) = 0$, that is, where $1+2\sin(x)=0$. Solving for $x$, we have $$\sin x = -\frac{1}{2};$$ in the interval in question, there are two points where $\sin(x)=-\frac{1}{2}$; they are $7\pi/6$ and $11\pi/6$. So these two points "break up" the interval $[0,2\pi]$ into three portions: $[0,7\pi/6]$, $[7\pi/6,11\pi/6]$, and $[11\pi/6,2\pi]$.

We then determine what sign the first derivative has on each of those intervals. On $(0,7\pi/6)$, the derivative is positive (we can ascertain this by plugging into the derivative any number between $0$ and $7\pi/6$ and seeing if we get something positive or something negative; $x=\pi$ is pretty easy, and $f'(\pi)=1\gt 0$). So $f(x)$ is increasing on $[0,7\pi/6]$.

On $(7\pi/6,11\pi/6)$, the first derivative is negative (if we plug in $x=3\pi/2$, we get $f'(3\pi/2) = 1 -2 = -1\lt 0$). So $f(x)$ is decreasing on $[7\pi/6,11\pi/6]$.

And on $(11\pi/6,2\pi)$, the derivative is positive (plugging in a value very close to $2\pi$ will give you $f'(x)$ very close to $1$); so $f(x)$ is increasing on $[11\pi/6,2\pi]$.

In summary, $f(x)$ is increasing on $[0,7\pi/6]\cup[11\pi/6,2\pi]$, and $f(x)$ is decreasing on $[7\pi/6,11\pi/6]$.

Looks like your first mistake was a sign error (you got $\pi/6$ instead of $7\pi/6$ for the critical point); then you forgot that there is a second point on $[0,2\pi]$ with the same sine value; and finally, you ignored the fact that the question restricted you to the interval $[0,2\pi]$, so there is no reason to be talking about any number less than $0$ or greater than $2\pi$.

For (3), the critical number cannot be $(2-x)/x^2$. That's also not the derivative or the second derivative.

Note that $f(x) = \frac{x-1}{x} = 1 - \frac{1}{x} = 1 - x^{-1}$. So \begin{align*} f'(x) &= -(-1)x^{-2} = x^{-2} = \frac{1}{x^2}\\ f''(x) &= (-2)x^{-3} = -\frac{2}{x^3}. \end{align*} If you used the quotient rule, you should have gotten: \begin{align*} f'(x) &= \frac{x(x-1)' - (x-1)(x)'}{x^2} = \frac{x-(x-1)}{x^2} = \frac{1}{x^2},\\ f''(x) &= \frac{x^2(1)' - (1)(x^2)'}{(x^2)^2} = \frac{0 - 2x}{x^4} = \frac{-2x}{x^4} = -\frac{2}{x^3}. \end{align*} So it seems you didn't do the derivatives correctly. Now, the function is concave up where $f''(x)\gt 0$, and concave down there $f''(x)\lt 0$. First we determine the points where $f''(x)$ can change signs, which are the points where $f''(x)$ is either undefined or equal to $0$.

$f''(x) = -\frac{2}{x^3}$ is never equal to zero; but it is undefined at $x=0$. So the only point where it can change signs is at $x=0$. This breaks up the real line into two parts: $(-\infty,0)$ and $(0,\infty)$.

On $(-\infty,0)$ the second derivative is positive (since $x$ is negative, so $x^3$ is negative, so $\frac{2}{x^3}$ is negative, so $-\frac{2}{x^3}$ is positive), so $f(x)$ is concave up on $(-\infty,0)$. On $(0,\infty)$, $f''(x)$ is negative, so $f(x)$ is concave down on $(0,\infty)$.

For the limit, it seems you are trying to use L'Hopital's Rule. First we need to write it as a quotient, check that we are in a situation where L'Hopital's Rule applies, and then apply L'Hopital's Rule. We have: \begin{align*} \lim_{x\to\pi/2}\left(\sec x - \tan x\right) &= \lim_{x\to\pi/2}\left(\frac{1}{\cos x} - \frac{\sin x}{\cos x}\right)\\ &= \lim_{x\to\pi/2}\frac{1 - \sin x}{\cos x}. \end{align*} When $x=\pi/2$, the numerator evaluates to $0$, as does the denominator. So we can use L'Hopital's Rule. We have: \begin{align*} \lim_{x\to\pi/2}\frac{1 - \sin x}{\cos x} &\stackrel{\mathrm{L'H}}{=} \lim_{x\to\pi/2}\frac{(1-\sin x)'}{(\cos x)'}\\ &= \lim_{x\to\pi/2}\frac{-\cos x}{-\sin x}\\ &= \lim_{x\to\pi/2}\frac{\cos x}{\sin x}\\ &= \frac{\cos(\pi/2)}{\sin(\pi/2)} = \frac{0}{1} = 0. \end{align*} So, it looks to me like that one is correct, modulo the fact that what you write is technically incorrect ($-\cos x/-\sin x$ is not equal to $0$, it is the limit as $x\to\pi/2$ that is equal to $0$).