# Is f(x)=cot(2x)*tanx^2 increasing or decreasing at x=pi/3?

Dec 14, 2017

$f \left(x\right)$ is decreasing at $x = \frac{\pi}{3}$.

#### Explanation:

$f \left(x\right) = \cot \left(2 x\right) \cdot \tan \left({x}^{2}\right)$

f'(x) = −2[csc^2(2x)tan(x^2)−xcot(2x)sec^2(x^2)]

Calculating value of $f ' \left(x\right)$ at $x = \frac{\pi}{3}$,

f'(pi/3) = −2[csc^2((2pi)/3)tan((pi/3)^2)−(pi/3)cot((2pi)/3)sec^2((pi/3)^2)]

f'(pi/3) = −2[csc^2((2pi)/3)tan(pi^2/9)−(pi/3)cot((2pi)/3)sec^2(pi^2/9)]

We know that,
$\csc \left(\frac{2 \pi}{3}\right) = 1.1547$, $\tan \left(\frac{2 \pi}{3}\right) = - 1.7320$, $\tan \left({\pi}^{2} / 9\right) = 1.948453$ and $\sec \left({\pi}^{2} / 9\right) = 2.19$

=>f'(pi/3) = −2[1.1547^2*1.948453−(-pi/(3xx1.7320))*2.19^2]

=>f'(pi/3) = −2[1.3333*1.948453+(pi/5.196)*4.7961]

=>f'(pi/3) = −2[2.5979+(15.06739/5.196)]

We can see that the slope $f ' \left(x\right)$ at $x = \frac{\pi}{3}$ is negative so the function $f \left(x\right)$ is decreasing at $x = \frac{\pi}{3}$.