What are the local extrema of #f(x)= sinx# on #[0,2pi]#?

Question

Christopher Agresta · Accepted Answer

At #x=pi/2# #f''(x)=-1# we have a local maxima and at #x=3pi/2#, #f''(x)=1# we have a local minima.

A maxima is a high point to which a function rises and then falls again. As such the slope of the tangent or the value of derivative at that point will be zero. Further, as the tangents to the left of maxima will be sloping upwards, then flattening and then sloping downwards, slope of the tangent will be continuously decreasing, i.e. the value of second derivative would be negative. A minima on the other hand is a low point to which a function falls and then rises again. As such the tangent or the value of derivative at minima too will be zero. But, as the tangents to the left of minima will be sloping downwards, then flattening and then sloping upwards, slope of the tangent will be continuously increasing or the value of second derivative would be positive. However, these maxima and minima may either be universal i.e. maxima or minima for the entire range or may be localized, i.e. maxima or minima in a limited range. Let us see this with reference to the function described in the question and for this let us first differentiate #f(x)=sinx#. #f'(x)=cosx# and on #[0,2pi]# it is #0# at #x=pi/2# and #x=(3pi)/2#. #f''(x)=-sinx# and while at #x=pi/2# #f''(x)=-1# meaning we have a local maxima, at #x=3pi/2#, #f''(x)=1# meaning we have a local minima. graph{sinx [-1, 7, -1.5, 1.5]}

Emily Ammerman · Answer

undefined The local extrema of $ f(x) = \sin(x) $ on the interval $[0, 2\pi]$ occur at critical points where the derivative equals zero or is undefined. Taking the derivative of $ f(x) $ with respect to $ x $ yields $ f'(x) = \cos(x) $. Setting $ f'(x) $ equal to zero gives us $ \cos(x) = 0 $. The solutions to this equation within the interval $[0, 2\pi]$ are $ x = \frac{\pi}{2} $ and $ x = \frac{3\pi}{2} $. At these critical points, $ f(x) $ reaches local extrema.

Evaluating $ f(x) $ at these critical points, we find $ f\left(\frac{\pi}{2}\right) = 1 $ and $ f\left(\frac{3\pi}{2}\right) = -1 $. Therefore, $ f(x) $ has a local maximum of $ 1 $ at $ x = \frac{\pi}{2} $ and a local minimum of $ -1 $ at $ x = \frac{3\pi}{2} $ on the interval $[0, 2\pi]$.