Given a function f(x) defined in the open interval (−2π,23π) as below,
f(x)=⎩⎨⎧2sin3xcosx(−2π<x<4π)(4π≤x<23π)
for a real number t, draw the graph g(t) which represents the number of real number k’s such that
(i) −2π<k<23π
(ii) ∣f(x)−t∣ is not differentiable atx=k
First, draw the graph of f(x). Although sinx is convex in (−2π,0) and concave in (0,2π), it is suspicious for 2sin3x to do so because ∣sinx∣≤1. To check the convexity of 2sin3x, the sign of its second-order function 6sinx(2cos2x−sin2x) could be used. Since the sign of f′′(x) is changed near x=0 from negative to positive, the convexity of 2sin3x is opposite to sinx. The graph of f(x) is as below. Note that f(x) has the local extrema (0,0) and (π,−1), other than the inflection point (0,0).
Now, to draw the graph of g(t), Knowing that the form of ∣f(x)−t∣ usually has sharp points at the folded points which are not differentiable, g(t) would be not continuous.
However, the unclear issue points are t=−1,0. Looking first t=0, it is unclear because f(x) has the inflection point at x=0. To check if ∣2sin3x∣ is differentiable at x=0, x→0+lim(2sin3x)′x→0−lim(−2sin3x)′=x→0+lim22sin3x6sin2xcosx=x→0+lim2sinxsinx3sin2xcosx=x→0+lim23sinxcosx=0=x→0−lim2−2sin3x−6sin2xcosx=x→0−lim2(−sinx)−sinx−3sin2xcosx=x→0−lim23−sinxcosx=0
So, ∣2sin3x∣ is differentiable at x=0 and g(0)=2. Now, looking into the case of t=−1, it is much harder to notice that ∣cosx+1∣ may not be differentiable since ∣cosx+1∣ is already smooth. But it could be. To check if ∣cosx+1∣ is differentiable at x=π, x→π+lim(cosx+1)′x→π−lim(cosx+1)′=x→π+lim2cosx+1−sinx=x→π+lim2cosx+11−cosx−sinx1−cosx=x→π+lim2sin2x−sinx1−cosx=x→π+lim−2sinx−sinx1−cosx=22=x→π−lim2cosx+1−sinx=x→π−lim2cosx+11−cosx−sinx1−cosx=x→π−lim2sin2x−sinx1−cosx=x→π−lim2sinx−sinx1−cosx=2−2
So, ∣cosx+1∣ is not differentiable at x=π and g(−1)=3.
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