Is $f (x) = cot (2 x) \cdot {tan x}^{2}$ $f(x)=cot(2x)*tanx^2$ increasing or decreasing at $x = \frac{π}{3}$ $x=pi/3$ ?

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Answer 1 · 2017-12-14T17:13:46

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

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

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

Calculating value of $f'(x)$ at $x= pi/3$ ,

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

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

We know that,
$csc((2pi)/3) = 1.1547$ , $tan((2pi)/3)=-1.7320$ , $tan(pi^2/9) = 1.948453$ and $sec(pi^2/9) = 2.19$

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

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

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

We can see that the slope $f'(x)$ at $x=pi/3$ is negative so the function $f(x)$ is decreasing at $x=pi/3$ .

Is f(x)=cot(2x)*tanx^2f(x)=cot(2x)⋅tanx2f(x)=cot(2x)*tanx^2 increasing or decreasing at x=pi/3x=π3x=pi/3?