006. Differentiability of Unknown Functions

Given a function f(x)f(x) defined in the open interval (π2,3π2)(-\frac{\pi}{2}, \frac{3\pi}{2}) as below,

f(x)={  2sin3x(π2<x<π4)  cosx(π4x<3π2)\begin{aligned} f(x) = \begin{cases} \; 2 \sin^3 x \quad &\left( -\cfrac{\pi}{2} < x < \cfrac{\pi}{4} \right) \\\\ \; \cos x \quad &\left( \cfrac{\pi}{4} \leq x < \cfrac{3\pi}{2} \right) \end{cases} \end{aligned}

for a real number tt, draw the graph g(t)g(t) which represents the number of real number kk’s such that

(i) π2<k<3π2-\cfrac{\pi}{2} < k < \cfrac{3\pi}{2}

(ii) f(x)t\sqrt{ \vert f(x) - t \vert } is not differentiable at x=kx = k

First, draw the graph of f(x)f(x). Although sinx\sin x is convex in (π2,0)(-\frac{\pi}{2}, 0) and concave in (0,π2)(0, \frac{\pi}{2}), it is suspicious for 2sin3x2 \sin^3 x to do so because sinx1\vert \sin x \vert \leq 1. To check the convexity of 2sin3x2 \sin^3 x, the sign of its second-order function 6sinx(23sin2x)6 \sin x (2 - 3 \sin^2 x) could be used. Since the sign of f(x)f''(x) is changed near x=0x = 0 from negative to positive, the convexity of 2sin3x2 \sin^3 x is opposite to sinx\sin x. Observing that f(x)f(x) has the inflection point, the graph of f(x)f(x) as below. Note that f(x)f(x) has another inflection point in (π2,0)(-\frac{\pi}{2}, 0).


Now, to draw the graph of g(t)g(t), Knowing that the form of f(x)t\vert f(x) - t \vert usually has sharp points at the folded points which are not differentiable, g(t)g(t) would be not continuous.


However, the unclear issue points are t=1,0t = -1, 0. Looking first t=0t = 0, it is unclear because f(x)f(x) has the inflection point at x=0x = 0. To check if 2sin3x\sqrt{\vert 2 \sin^3 x \vert} is differentiable at x=0x = 0, limx0+(2sin3x)=limx0+6sin2xcosx22sin3x=limx0+3sin2xcosx2sinxsinx=limx0+3sinxcosx2=0limx0(2sin3x)=limx06sin2xcosx22sin3x=limx03sin2xcosx2(sinx)sinx=limx03sinxcosx2=0\begin{aligned} \lim_{x \to 0+} \left( \sqrt{2 \sin^3 x} \right)' &= \lim_{x \to 0+}\cfrac{6 \sin^2 x \cos x}{2 \sqrt{2 \sin^3 x}} = \lim_{x \to 0+} \cfrac{3 \sin^2 x \cos x}{\sqrt{2} \sin x \sqrt{\sin x}} = \lim_{x \to 0+} \cfrac{3 \sqrt{\sin x} \cos x}{\sqrt{2}} = 0 \\\\ \lim_{x \to 0-} \left( \sqrt{-2 \sin^3 x} \right)' &= \lim_{x \to 0-} \cfrac{-6 \sin^2 x \cos x}{2 \sqrt{-2 \sin^3 x}} = \lim_{x \to 0-} \cfrac{-3 \sin^2 x \cos x}{\sqrt{2} (-\sin x) \sqrt{-\sin x}} = \lim_{x \to 0-} \cfrac{3 \sqrt{-\sin x} \cos x}{\sqrt{2}} = 0 \end{aligned}

So, 2sin3x\sqrt{\vert 2 \sin^3 x \vert} is differentiable at x=0x = 0 and g(0)=2g(0) = 2. Now, looking into the case of t=1t = -1, it is much harder to notice that cosx+1\sqrt{ \vert \cos x + 1 \vert } may not be differentiable since cosx+1\vert \cos x + 1 \vert is already smooth. But it could be. To check if cosx+1\sqrt{\vert \cos x + 1 \vert} is differentiable at x=πx = \pi, limxπ+(cosx+1)=limxπ+sinx2cosx+1=limxπ+sinx1cosx2cosx+11cosx=limxπ+sinx1cosx2sin2x=limxπ+sinx1cosx2sinx=22limxπ(cosx+1)=limxπsinx2cosx+1=limxπsinx1cosx2cosx+11cosx=limxπsinx1cosx2sin2x=limxπsinx1cosx2sinx=22\begin{aligned} \lim_{x \to \pi+} \left( \sqrt{\cos x + 1} \right)' &= \lim_{x \to \pi+} \cfrac{-\sin x}{2 \sqrt{\cos x + 1}} = \lim_{x \to \pi+} \cfrac{-\sin x \sqrt{1 - \cos x}}{2 \sqrt{\cos x + 1}\sqrt{1 - \cos x}}\\\\ &= \lim_{x \to \pi+} \cfrac{-\sin x \sqrt{1 - \cos x}}{2 \sqrt{\sin^2 x}} = \lim_{x \to \pi+} \cfrac{-\sin x \sqrt{1 - \cos x}}{-2 \sin x} = \cfrac{\sqrt{2}}{2} \\\\ \lim_{x \to \pi-} \left( \sqrt{\cos x + 1} \right)' &= \lim_{x \to \pi-} \cfrac{-\sin x}{2 \sqrt{\cos x + 1}} = \lim_{x \to \pi-} \cfrac{-\sin x \sqrt{1 - \cos x}}{2 \sqrt{\cos x + 1}\sqrt{1 - \cos x}}\\\\ &= \lim_{x \to \pi-} \cfrac{-\sin x \sqrt{1 - \cos x}}{2 \sqrt{\sin^2 x}} = \lim_{x \to \pi-} \cfrac{-\sin x \sqrt{1 - \cos x}}{2 \sin x} = \cfrac{-\sqrt{2}}{2} \end{aligned}

So, cosx+1\sqrt{\vert \cos x + 1\vert} is not differentiable at x=πx = \pi and g(1)=3g(-1) = 3.


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