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Let \(f(x)=\cos(2x)+2\cos(x)\).
a) Show that \(f(x)=2\left(\cos(x)+\frac{1}{2}\right)^2-\frac{3}{2}\).
b) Determine whether the graph has y-axis symmetry.
c) Use the form from part a) to find the global maximum and minimum values of \(f\) on \(\mathbb{R}\). Justify your answer.
Hints
- Use the double-angle identity for cosine.
- Recall the condition for an even function.
- Treat \(\cos(x)\) as a variable restricted to \([-1,1]\).
- Compare the vertex value with the values at both endpoints of that interval.
Solution
1. Use \(\cos(2x)=2\cos^2(x)-1\): \(f(x)=2\cos^2(x)+2\cos(x)-1\). Completing the square gives \(f(x)=2\left(\cos(x)+\frac{1}{2}\right)^2-\frac{3}{2}\).
2. Since cosine is even, \(f(-x)=\cos(-2x)+2\cos(-x)=f(x)\). Therefore, the graph has y-axis symmetry.
3. Let \(u=\cos(x)\), where \(-1\le u\le1\). Then \(f=2\left(u+\frac{1}{2}\right)^2-\frac{3}{2}\). The minimum occurs at \(u=-\frac{1}{2}\), giving \(-\frac{3}{2}\). For the maximum, compare the endpoints: at \(u=-1\), the value is \(-1\), and at \(u=1\), the value is \(3\). Thus the global maximum is \(3\).
Answer
a) \(f(x)=2\left(\cos(x)+\frac{1}{2}\right)^2-\frac{3}{2}\).
b) Yes. Since \(f(-x)=f(x)\), the graph has y-axis symmetry.
c) Global minimum: \(-\frac{3}{2}\). Global maximum: \(3\).
