# 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.

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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 $0$).

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