Is f(x)=e^-xcos(-x)-sinx/(pi-e^x) increasing or decreasing at x=pi/6?

1 Answer

f(x) is decreasing at x=\pi/6

Explanation:

Given function:

f(x)=e^{-x}\cos(-x)-\frac{\sin x}{\pi-e^x}

f(x)=e^{-x}\cosx+\frac{\sin x}{e^x-\pi}

Differentiating above function w.r.t. x using chain rule & quotient rule as follows

f'(x)=d/dx(e^{-x}\cosx+\frac{\sin x}{e^x-\pi})

=e^{-x}(-\sinx)+\cos x(-e^{-x})+\frac{(e^x-\pi)\cos x-\sin x(e^x)}{(e^x-\pi)^2}

=-e^{-x}(\sinx+\cos x)+\frac{e^x(\cos x-\sin x)-\pi\cos x}{(e^x-\pi)^2}

Setting x=\pi/6 in above expression we get

f'(\pi/6)

=-e^{-\pi/6}(\sin(\pi/6)+\cos (\pi/6))+\frac{e^{\pi/6}(\cos (\pi/6)-\sin (\pi/6))-\pi\cos (\pi/6)}{(e^{\pi/6}-\pi)^2}

=-0.80921-0.9953

=-1.8045

Since f'(\pi/6)<0 hence the given function f(x) is decreasing at x=\pi/6