Question

Rolle's theorem holds for monic quadratic polynomial \(f(x)\) on the interval \([α,α+3]\) where \(f(α) = 0\). Similarly, \(g(x) = f(x)+2\) also follows Rolle's theorem in the interval \([β,3]\) where \(g(3) = 0\), such that the value of \(c\) is the same for both \(f(x)\) and \(g(x)\). Then the value of \((f\circ g)(α)\) is...

• A. \(-4\)
• B. \(4\)
• C. \(-2\)
• D. \(2\)

Hint:

Rolle needs equal end values, so a monic quadratic has its vertex at the point c.

Answer 1

The Correct Option is C

Step 1: Rolle for f
\(f\) is monic quadratic with \(f(\alpha) = 0\) and Rolle on \([\alpha, \alpha+3]\) needs \(f(\alpha + 3) = 0\). So \(f(x) = (x - \alpha)(x - \alpha - 3)\) and \(c = \alpha + \frac{3}{2}\) (the vertex).

Step 2: Rolle for g
\(g(x) = f(x) + 2\) has the same vertex, so \(c\) is the same. Write \(g(x) = (x - c)^2 - \frac{1}{4}\), since \(f(x) = (x-c)^2 - \frac{9}{4}\).

Step 3: Use g(3) = 0
\((3 - c)^2 = \frac{1}{4}\) so \(c = \frac{5}{2}\) or \(c = \frac{7}{2}\). Also \(g(\beta) = 0\) gives \(\beta = 2c - 3\), and the interval \([\beta, 3]\) needs \(\beta < 3\), so \(c = \frac{5}{2}\), \(\beta = 2\), and \(\alpha = c - \frac{3}{2} = 1\).

Step 4: Compute
\(g(\alpha) = g(1) = \left(1 - \frac{5}{2}\right)^2 - \frac{1}{4} = \frac{9}{4} - \frac{1}{4} = 2\). Then \(f(2) = (2 - 1)(2 - 4) = -2\). Option (C).

Final Answer:
(f o g)(alpha) equals -2. This is option (C). \[ \boxed{\text{(C) }-2} \]