Is #f(x)=sinx/x# increasing or decreasing at #x=pi/3#?

Question

Christopher Allers · Accepted Answer

Decreasing.

To determine if a function is increasing or decreasing at a point, use the function's derivative: So, we first must find the derivative of #f#. To do so, we will have to use the quotient rule. Application of the quotient rule shows that #f'(x)=(xd/dx(sinx)-sinxd/dx(x))/x^2# #=(xcosx-sinx)/x^2# So, to determine if #f# is increasing or decreasing at #x=pi/3#, find #f'(pi/3)# and see if it is positive or negative. #f'(pi/3)=(pi/3cos(pi/3)-sin(pi/3))/(pi/3)^2# #=(pi/3(1/2)-sqrt3/2)/(pi^2/9)approx-0.3123# Since this is #<0#, the function is decreasing at #x=pi/3#. We can check a graph of #f# #(#note that #pi/3approx1.0472)#. graph{sinx/x [-3.945, 4.825, -1.568, 2.817]}

Aurora Alvarado · Answer

undefined To determine whether $ f(x) = \frac{\sin x}{x} $ is increasing or decreasing at $ x = \frac{\pi}{3} $, we need to analyze the sign of its derivative at that point. Let's find the derivative of $ f(x) $ with respect to $ x $ using the quotient rule:

$$ f'(x) = \frac{(x)(\cos x) - (\sin x)(1)}{x^2} $$

Simplify this expression:

$$ f'(x) = \frac{x \cdot \cos x - \sin x}{x^2} $$

Now, evaluate $ f'(x) $ at $ x = \frac{\pi}{3} $:

$$ f'\left(\frac{\pi}{3}\right) = \frac{\frac{\pi}{3} \cdot \cos\left(\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{3}\right)}{\left(\frac{\pi}{3}\right)^2} $$

$$ f'\left(\frac{\pi}{3}\right) = \frac{\frac{\pi}{3} \cdot \frac{1}{2} - \frac{\sqrt{3}}{2}}{\left(\frac{\pi}{3}\right)^2} $$

$$ f'\left(\frac{\pi}{3}\right) = \frac{\frac{\pi}{6} - \frac{\sqrt{3}}{2}}{\left(\frac{\pi}{3}\right)^2} $$

$$ f'\left(\frac{\pi}{3}\right) = \frac{\pi - 3\sqrt{3}}{2\pi^2} $$

Since $ \pi - 3\sqrt{3} $ is positive and $ 2\pi^2 $ is positive, $ f'\left(\frac{\pi}{3}\right) $ is positive.

Therefore, $ f(x) = \frac{\sin x}{x} $ is increasing at $ x = \frac{\pi}{3} $.