# Is f(x)= cot(-x+(5pi)/6)  increasing or decreasing at x=pi/4 ?

Increasing

#### Explanation:

Given function:

$f \left(x\right) = \setminus \cot \left(- x + \frac{5 \setminus \pi}{6}\right)$

Differentiating above equation w.r.t. $x$ as follows

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} \setminus \cot \left(- x + \frac{5 \setminus \pi}{6}\right)$

$f ' \left(x\right) = - \setminus \cos e {c}^{2} \left(- x + \frac{5 \setminus \pi}{6}\right) \frac{d}{\mathrm{dx}} \left(- x + \frac{5 \setminus \pi}{6}\right)$

$f ' \left(x\right) = - \setminus \cos e {c}^{2} \left(- x + \frac{5 \setminus \pi}{6}\right) \left(- 1\right)$

$f ' \left(x\right) = \setminus \cos e {c}^{2} \left(- x + \frac{5 \setminus \pi}{6}\right)$

setting $x = \setminus \frac{\pi}{4}$ in above equation, we get

$f ' \left(\setminus \frac{\pi}{4}\right) = \setminus \cos e {c}^{2} \left(- \setminus \frac{\pi}{4} + \frac{5 \setminus \pi}{6}\right)$

$= \setminus \cos e {c}^{2} \left(\frac{7 \setminus \pi}{12}\right)$

$= \setminus \cos e {c}^{2} {105}^{\setminus} \circ$

$= {\left(\setminus \frac{2 \setminus \sqrt{2}}{\setminus \sqrt{3} + 1}\right)}^{2}$

$= 4 \left(2 - \setminus \sqrt{3}\right) > 0$

Since, $f ' \left(\setminus \frac{\pi}{4}\right) > 0$ hence the function is increasing at $x = \setminus \frac{\pi}{4}$

Jul 18, 2018

since we get $f ' \left(\frac{\pi}{4}\right) = 2 {\left(\sqrt{3} - 1\right)}^{2} > 0$ so is $f \left(x\right)$ increasing for this Point.

#### Explanation:

At first we not that

$\cot \left(- x + 5 \frac{\pi}{6}\right) = - \cot \left(x + \frac{\pi}{6}\right)$
so

$- \cot \left(x + \frac{\pi}{6}\right) = {\csc}^{2} \left(x + \frac{\pi}{6}\right)$
and

$f ' \left(\frac{\pi}{4}\right) = 2 {\left(\sqrt{3} - 1\right)}^{2}$
Note that $\frac{\pi}{4} + \frac{\pi}{6} = \frac{3 \pi + 2 \pi}{12} = \frac{5 \pi}{12}$