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Problem #549
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549
The function \(f : x \mapsto 5 + 3 \cos\left(\frac{1}{2}x\right)\) is defined for \(0 \leq x \leq 2\pi\).
Solve the equation \(f(x) = 7\), giving your answer correct to 2 decimal places. [3]
Sketch the graph of \(y = f(x)\). [2]
Explain why \(f\) has an inverse. [1]
Obtain an expression for \(f^{-1}(x)\). [3]
Solution
(i) Start with the equation \(5 + 3 \cos\left(\frac{1}{2}x\right) = 7\). Subtract 5 from both sides to get \(3 \cos\left(\frac{1}{2}x\right) = 2\). Divide by 3: \(\cos\left(\frac{1}{2}x\right) = \frac{2}{3}\). Find \(\frac{1}{2}x = \cos^{-1}\left(\frac{2}{3}\right) \approx 0.84\). Multiply by 2 to solve for \(x\): \(x \approx 1.68\).
(ii) The graph of \(y = 5 + 3 \cos\left(\frac{1}{2}x\right)\) is a cosine wave starting at \(y = 8\) when \(x = 0\) and ending at \(y = 2\) when \(x = 2\pi\).
(iii) The function \(f\) has no turning points and is one-to-one over the given interval, so it has an inverse.
(iv) To find \(f^{-1}(x)\), start with \(y = 5 + 3 \cos\left(\frac{1}{2}x\right)\). Rearrange to make \(x\) the subject: \(y - 5 = 3 \cos\left(\frac{1}{2}x\right)\), then \(\cos\left(\frac{1}{2}x\right) = \frac{y-5}{3}\). Solve for \(x\): \(\frac{1}{2}x = \cos^{-1}\left(\frac{y-5}{3}\right)\), so \(x = 2 \cos^{-1}\left(\frac{y-5}{3}\right)\).
