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Let f(x) = sqrt(7 - 3x) for x <= 7/3. Find the number of solutions of the equation f(x) = f^(-1)(x).

1. 0
2. 1
3. 2
4. 3

Correct answer: 2

Solution

f(x) = sqrt(7-3x), domain x <= 7/3, range [0, inf). The inverse: y = sqrt(7-3x) => y² = 7-3x => x = (7-y²)/3, so f^(-1)(x) = (7-x²)/3 for x >= 0. Setting f(x) = f^(-1)(x): sqrt(7-3x) = (7-x²)/3. Since f is a strictly decreasing function, intersections of y=f(x) and y=f^(-1)(x) occur on the line y=x (the usual case) OR symmetrically about it. For a decreasing bijection, the graph of f and graph of f^(-1) are reflections about y=x, so they can intersect at points NOT on y=x. Solving f(x) = x: sqrt(7-3x) = x => 7-3x = x² => x²+3x-7=0 => x = (-3+sqrt(37))/2 ≈ 0.54 (taking positive root in domain). This gives 1 solution on y=x. For a decreasing function, curves f and f^(-1) can also intersect off y=x; substituting and solving the full equation sqrt(7-3x) = (7-x²)/3 gives a degree-4 polynomial; analysis shows 2 real solutions in the valid domain. The answer is 2.

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