Porisma

Worked solution

A piecewise inverse and two identity domains

The function ff is defined by

f:x↦{cos⁡(πx/2),0≤x<1,1−x,1≤x≤2.f:x\mapsto\begin{cases}\cos(\pi x/2),&0\leq x<1,\\1-x,&1\leq x\leq2.\end{cases}

(i) Sketch the graph of ff, indicating clearly the coordinates of the axial intercepts and endpoints.

(ii) Show that ff has an inverse, and define it in similar form.

(iii) Find the range of values of xx for which ff−1(x)=f−1f(x)ff^{-1}(x)=f^{-1}f(x).

(iv) The function gg is defined by g:x↦exg:x\mapsto e^x, 0≤x≤ln⁡20\leq x\leq\ln2.

Find the range of fgfg.

[2020/HCI/Promo/11]

CJC · Functions · Discussion 6 · 2025

Answer

(ii) f−1(x)=(2/π)cos⁡−1xf^{-1}(x)=(2/\pi)\cos^{-1}x for 0<x≤10<x\leq1, and 1−x1-x for −1≤x≤0-1\leq x\leq0. (iii) [0,1][0,1]. (iv) [−1,0][-1,0].

(i)

The cosine branch decreases from closed (0,1) towards (1,0). The line branch includes (1,0) and ends at closed (2,−1). Thus the y-intercept is (0,1), the x-intercept is (1,0), and both outer endpoints are included.

How did this part go?

(ii)

Every horizontal line meets the graph at most once: f is strictly decreasing over [0,2], so it is one-to-one. On 0≤x<1, solve y=cos⁡(πx/2)y=\cos(\pi x/2) to obtain x=(2/π)cos⁡−1yx=(2/\pi)\cos^{-1}y. On 1≤x≤2, solve y=1−x to get x=1−y. Therefore f−1:x↦{(2/π)cos⁡−1x,0<x≤1,1−x,−1≤x≤0.f^{-1}:x\mapsto\begin{cases}(2/\pi)\cos^{-1}x,&0<x\leq1,\\1-x,&-1\leq x\leq0.\end{cases} The inverse domain is [−1,1]. As the source notes, x=0 can be attached to either branch or both because both formulas give 1 there.

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(iii)

Both composites have identity rule x, but Dff−1=Df−1=[−1,1]D_{ff^{-1}}=D_{f^{-1}}=[-1,1] and Df−1f=Df=[0,2]D_{f^{-1}f}=D_f=[0,2]. They can be compared only on the intersection, [0,1]. Hence 0≤x≤10\leq x\leq1.

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(iv)

The graph of g has closed endpoints (0,1) and (ln2,2). It maps Dg=[0,ln⁡2]D_g=[0,\ln2] to Rg=[1,2]R_g=[1,2], where the outer f uses the linear branch 1−x. The mapping chain is [0,ln⁡2]→g[1,2]→f[−1,0][0,\ln2]\xrightarrow{g}[1,2]\xrightarrow{f}[-1,0]. Thus Rfg=[−1,0]R_{fg}=[-1,0].

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Compare your sketch

Expand diagram

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The first graph decreases from closed (0,1) along cos(πx/2) to (1,0), which is included by the line branch 1−x; that line ends at closed (2,−1). The second graph is eˣ on [0,ln2] with closed endpoints (0,1) and (ln2,2). The source mapping chain is [0,ln2] through g to [1,2], then through f to [−1,0].
The first graph decreases from closed (0,1) along cos(πx/2) to (1,0), which is included by the line branch 1−x; that line ends at closed (2,−1). The second graph is eˣ on [0,ln2] with closed endpoints (0,1) and (ln2,2). The source mapping chain is [0,ln2] through g to [1,2], then through f to [−1,0].
The first graph decreases from closed (0,1) along cos(πx/2) to (1,0), which is included by the line branch 1−x; that line ends at closed (2,−1). The second graph is eˣ on [0,ln2] with closed endpoints (0,1) and (ln2,2). The source mapping chain is [0,ln2] through g to [1,2], then through f to [−1,0].

Common mistakes

Having the same identity rule does not make the two composites equal on different domains.

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