Is $f (x) = cot (- x + \frac{5 π}{6})$ $f(x)= cot(-x+(5pi)/6)$ increasing or decreasing at $x = \frac{π}{4}$ $x=pi/4$ ?

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Answer 1 · 2018-07-18T17:38:53

Increasing

Given function:

$f (x) = cot (- x + \frac{5 π}{6})$ $f(x)=\cot(-x+{5\pi}/6)$

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

$\frac{d}{d x} f (x) = \frac{d}{d x} cot (- x + \frac{5 π}{6})$ $d/dxf(x)=d/dx\cot(-x+{5\pi}/6)$

$f'(x)=-\cosec^2(-x+{5\pi}/6)d/dx(-x+{5\pi}/6)$

$f'(x)=-\cosec^2(-x+{5\pi}/6)(-1)$

$f'(x)=\cosec^2(-x+{5\pi}/6)$

setting $x=\pi/4$ in above equation, we get

$f'(\pi/4)=\cosec^2(-\pi/4+{5\pi}/6)$

$=\cosec^2({7\pi}/12)$

$=\cosec^2 105^\circ$

$=(\frac{2\sqrt2}{\sqrt3+1})^2$

$=4(2-\sqrt3)>0$

Since, $f'(\pi/4)>0$ hence the function is increasing at $x=\pi/4$

Answer 2 · 2018-07-18T17:50:50

since we get $f'(pi/4)=2(sqrt(3)-1)^2>0$ so is $f(x)$ increasing for this Point.

At first we not that

$cot(-x+5pi/6)=-cot(x+pi/6)$
so

$-cot(x+pi/6)=csc^2(x+pi/6)$
and

$f'(pi/4)=2(sqrt(3)-1)^2$
Note that $pi/4+pi/6=(3pi+2pi)/12=(5pi)/12$

Is f(x)= cot(-x+(5pi)/6) f(x)=cot(−x+5π6)f(x)= cot(-x+(5pi)/6) increasing or decreasing at x=pi/4 x=π4x=pi/4 ?