The function g\text{g} is defined by g:x→sin⁡x+x\text{g} : x \to \sin x + x, for x∈Rx \in \mathbb{R}, −π≤x≤π-\pi \le x \le \pi.

  1. (i)

    Show that g′(x)>0\text{g}'(x) > 0 for −π<x<π-\pi < x < \pi. Hence, or otherwise, show that g\text{g} has an inverse.

  2. (ii)

    Sketch on the same diagram the graphs of y=g(x)y = \text{g}(x) and y=g−1(x)y = \text{g}^{-1}(x).

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Worked solution

2(i)

1. Find the first derivative of $\text{g}(x)$.

g′(x)=ddx(sin⁡x+x)=cos⁡x+1\text{g}'(x) = \frac{\text{d}}{\text{d}x}(\sin x + x) = \cos x + 1

2. Show that the derivative is strictly positive on the given open interval.

For −π<x<π-\pi < x < \pi, the range of cos⁡x\cos x is −1<cos⁡x≤1-1 < \cos x \le 1.
Since cos⁡x>−1\cos x > -1, it follows that cos⁡x+1>0\cos x + 1 > 0.
Hence, g′(x)>0\text{g}'(x) > 0 for −π<x<π-\pi < x < \pi.

3. Deduce that the function has an inverse.

Since g′(x)>0\text{g}'(x) > 0 for the interior of its domain, g\text{g} is a strictly increasing function.
A strictly increasing function is a one-to-one function (it passes the horizontal line test).
Therefore, the inverse function g−1\text{g}^{-1} exists.

Answer

Shown.

2(ii)

1. Identify key coordinates, relative position to $y=x$, and endpoint tangents of $y = \text{g}(x)$.

g(−π)=−π,g(0)=0,g(π)=π\text{g}(-\pi) = -\pi, \quad \text{g}(0) = 0, \quad \text{g}(\pi) = \pi
Since g′(±π)=cos⁡(±π)+1=0\text{g}'(\pm\pi) = \cos(\pm\pi) + 1 = 0, the graph of g\text{g} has horizontal tangents at the endpoints (−π,−π)(-\pi, -\pi) and (π,π)(\pi, \pi).
Furthermore, for 0<x<π0 < x < \pi, sin⁡x>0  ⟹  g(x)>x\sin x > 0 \implies \text{g}(x) > x, so the curve lies above the line y=xy=x. Conversely, for −π<x<0-\pi < x < 0, sin⁡x<0  ⟹  g(x)<x\sin x < 0 \implies \text{g}(x) < x, meaning the curve lies below y=xy=x.

2. Deduce the corresponding properties for $y = \text{g}^{-1}(x)$ using the reflection property of inverse functions.

The graph of y=g−1(x)y = \text{g}^{-1}(x) is the reflection of y=g(x)y = \text{g}(x) in the line y=xy = x.
Therefore, it will share the intersection points (−π,−π)(-\pi, -\pi), (0,0)(0,0), and (π,π)(\pi, \pi).
The horizontal tangents of g(x)\text{g}(x) become vertical tangents for g−1(x)\text{g}^{-1}(x) at the endpoints.
For 0<x<π0 < x < \pi, g−1(x)\text{g}^{-1}(x) lies below y=xy=x, and for −π<x<0-\pi < x < 0, it lies above y=xy=x.

3. Produce the combined sketch.

(Student draws the axes, sketching the line y=xy=x as a guide, and carefully draws y=g(x)y = \text{g}(x) and y=g−1(x)y = \text{g}^{-1}(x) exhibiting the correct intersection points, relative positions, and tangent behaviors at the extremities.)

Answer

A sketch containing both curves on the interval [−π,π][-\pi, \pi]. They intersect at (−π,−π)(-\pi, -\pi), (0,0)(0, 0), and (π,π)(\pi, \pi).

  • y=g(x)y = \text{g}(x) has horizontal tangents at the endpoints and lies above y=xy=x for 0<x<π0 < x < \pi.
  • y=g−1(x)y = \text{g}^{-1}(x) has vertical tangents at the endpoints and lies below y=xy=x for 0<x<π0 < x < \pi.

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