Question

for the function f(x)=csc x + cot x on the interval (0, 2pi)
a) determine where it is discontinuous. Identify any vertical asymptotes
b) determine the intervals of concavity and inflection points

Answer 1

For the function $f(x) = \csc x + \cot x$ on the interval $[0, 2pi]$

a) determine where it is discontinuous. Identify any vertical asymptotes

b) determine the intervals of concavity and inflection points

Solution.

a) First of all, let's transform our expression:

$$\csc x + \cot x = \frac{1}{\sin x} + \frac{\cos x}{\sin x} = \frac{1 + \cos x}{\sin x}$$

We can see that our function defined on the interval $[0, 2\pi]$ for all $x$ except $\sin x \neq 0 \Leftrightarrow x \neq 0, 2\pi$, where it has a discontinuity.

So let's find one-sided limits at

$$\lim_{x \to 0 \sim 0} \frac{1 + \cos x}{\sin x} = \left[ \frac{2}{-0} \right] = -\infty$$ $$\lim_{x \to 0 \pm 0} \frac{1 + \cos x}{\sin x} = \left[ \frac{2}{+0} \right] = +\infty$$ $$\lim_{x \to 2\pi \sim 0} \frac{1 + \cos x}{\sin x} = \left[ \frac{2}{-0} \right] = -\infty$$ $$\lim_{x \to 2\pi \pm 0} \frac{1 + \cos x}{\sin x} = \left[ \frac{2}{+0} \right] = +\infty$$

The function $f(x)$ have a discontinuity of the second kind at $x = 0, 2\pi$, because the one-sided limits are infinite.

Vertical asymptotes are vertical lines which correspond to the zeroes of the denominator of a rational function.

So on the interval $[0, 2\pi]$ the denominator has zeros at $x = 0, 2\pi \Rightarrow$

$\Rightarrow$ we have two vertical asymptotes: $x = 0, x = 2\pi$

b) we find the intervals of concavity and inflection points using the second derivative of the function $f(x) = \csc x + \cot x$.

$$f'(x) = (\csc x)' + (\cot x)' = -\cot x \csc x - \csc^2 x = -\csc x (\cot x + \csc x)$$ $$\begin{aligned} f'' &= \left(-\csc x (\cot x + \csc x)\right)' = -(\csc x)' (\cot x + \csc x) - \csc x (\cot x + \csc x)' \\ &= \cot x \csc x (\cot x + \csc x) + \csc^2 x (\cot x + \csc x) = \csc x (\cot x + \csc x) (\cot x + \csc x) \\ &= \csc x (\cot x + \csc x)^2 = \csc x \left(\frac{1 + \cos x}{\sin x}\right)^2 = \csc^3 x (1 + \cos x)^2 \\ f'' &= 0 \\ \csc^3 x (1 + \cos x)^2 &= 0 \\ \csc x &= 0 \Longrightarrow \text{no solutions} \end{aligned}$$ $$(1 + \cos x) ^ {2} = 0$$ $$1 + \cos x = 0$$ $$\cos x = - 1$$ $$x = \pi \text{ on the interval } x \in [0, 2\pi]$$ $$x = \pi - \text{inflection point}$$

On the interval $(0,\pi)$ $\cos x > 0 \Rightarrow$ concave up

On the interval $(\pi, 2\pi)$ $\cos x < 0 \Rightarrow$ concave down

Answer: $f(x)$ have a discontinuity at $x = 0, 2\pi$;

$f(x)$ have two vertical asymptotes: $x = 0, x = 2\pi$;

$f(x)$ concave up on the interval $(0,\pi)$;

$f(x)$ concave down on the interval $(\pi, 2\pi)$;

$f(x)$ have inflection point: $x = \pi$