Functions f and g are defined by
\(f(x) = (x + a)^2 - a\) for \(x \leq -a\),
\(g(x) = 2x - 1\) for \(x \in \mathbb{R}\),
where \(a\) is a positive constant.
Given that \(a = \frac{7}{2}\), solve the equation \(gf(x) = 0\).
Solution
Given \(a = \frac{7}{2}\), we need to solve \(gf(x) = 0\).
First, find \(f(x)\):
\(f(x) = \left(x + \frac{7}{2}\right)^2 - \frac{7}{2}\)
For \(x \leq -\frac{7}{2}\), \(f(x) = \left(x + \frac{7}{2}\right)^2 - \frac{7}{2}\).
Now, find \(gf(x)\):
\(gf(x) = g(f(x)) = 2f(x) - 1\)
Set \(gf(x) = 0\):
\(2f(x) - 1 = 0\)
\(2\left(\left(x + \frac{7}{2}\right)^2 - \frac{7}{2}\right) - 1 = 0\)
\(2\left(x + \frac{7}{2}\right)^2 - 7 - 1 = 0\)
\(2\left(x + \frac{7}{2}\right)^2 = 8\)
\(\left(x + \frac{7}{2}\right)^2 = 4\)
\(x + \frac{7}{2} = \pm 2\)
\(x = -\frac{7}{2} \pm 2\)
\(x = -\frac{7}{2} + 2 = -\frac{3}{2}\)
\(x = -\frac{7}{2} - 2 = -\frac{11}{2}\)
Since \(x \leq -\frac{7}{2}\), the valid solution is \(x = -\frac{11}{2}\).
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