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Problem 546
546
The function \(f\) is defined by \(f(x) = 3 \tan\left(\frac{1}{2}x\right) - 2\), for \(-\frac{1}{2}\pi \leq x \leq \frac{1}{2}\pi\).
(i) Solve the equation \(f(x) + 4 = 0\), giving your answer correct to 1 decimal place. [3]
(ii) Find an expression for \(f^{-1}(x)\) and find the domain of \(f^{-1}\). [5]
(iii) Sketch, on the same diagram, the graphs of \(y = f(x)\) and \(y = f^{-1}(x)\). [3]
Solution
(i) Start with the equation \(f(x) + 4 = 0\), which simplifies to \(3 \tan\left(\frac{1}{2}x\right) - 2 + 4 = 0\) or \(3 \tan\left(\frac{1}{2}x\right) = -2\). This gives \(\tan\left(\frac{1}{2}x\right) = -\frac{2}{3}\). Solving for \(\frac{1}{2}x\), we find \(\frac{1}{2}x = \tan^{-1}\left(-\frac{2}{3}\right) \approx -0.588\). Therefore, \(x = 2(-0.588) = -1.2\).
(ii) To find \(f^{-1}(x)\), start with \(y = 3 \tan\left(\frac{1}{2}x\right) - 2\). Rearrange to isolate \(\tan\left(\frac{1}{2}x\right)\): \(y + 2 = 3 \tan\left(\frac{1}{2}x\right)\), so \(\tan\left(\frac{1}{2}x\right) = \frac{y+2}{3}\). Taking the inverse tangent gives \(\frac{1}{2}x = \tan^{-1}\left(\frac{y+2}{3}\right)\), thus \(x = 2 \tan^{-1}\left(\frac{y+2}{3}\right)\). Therefore, \(f^{-1}(x) = 2 \tan^{-1}\left(\frac{x+2}{3}\right)\). The domain of \(f^{-1}\) is \(-5, 1\) based on the range of \(f\).
(iii) The sketch should show a tan graph through the first, third, and fourth quadrants, and an inverse tan graph through the first, second, and third quadrants. The two curves should be symmetrical about \(y = x\) and approximately in the correct domain and range.
