Name: 14. For this problem, \( f(x)=x^{2} \). Because \( f \) is differentiable everywhere, it satisfies all hypotheses of the MVT on any interval \( [a, b] \). a) What does the MVT say about \( f \) on \( [\mathrm{a}, \mathrm{b}]=[-1,1] \) ? Find all suitable values of \( c \). How many are there? b) What does the MVT say about \( f \) on \( [\mathrm{a}, \mathrm{b}]=[1,2] \) ? Find all suitable values of \( c \). How many are there? c) Show that for \( f(x)=x^{2} \) and any interval \( [a, b] \), the MVT's number \( c \) is the midpoint of \( [\mathrm{a}, \mathrm{b}] \).

The Mean Value Theorem (MVT) states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \(c\) in \((a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). For \(f(x) = x^2\), you can calculate its derivative \(f'(x) = 2x\) and apply the MVT to find those values of \(c\). For the interval \([-1, 1]\), \(f(-1) = 1\), \(f(1) = 1\), therefore \(\frac{f(1) - f(-1)}{1 - (-1)} = 0\). Setting \(f'(c) = 2c = 0\) gives \(c = 0\). Thus, there is **1 suitable value of \(c\)**, which is \(0\). For the interval \([1, 2]\), \(f(1) = 1\) and \(f(2) = 4\), thus \(\frac{f(2) - f(1)}{2 - 1} = 3\). Solving \(2c = 3\) results in \(c = \frac{3}{2}\), meaning there is **1 suitable value of \(c\)** here too, which is \(\frac{3}{2}\). To prove that \(c\) is always the midpoint of \([a, b]\), consider \(f'(c) = \frac{f(b) - f(a)}{b - a} = \frac{b^2 - a^2}{b - a} = \frac{(b + a)(b - a)}{b - a} = b + a\). Since \(f'(c) = 2c\), setting \(2c = b + a\) yields \(c = \frac{a + b}{2}\), confirming that \(c\) is indeed the midpoint of any interval \([a, b]\).