Question

let $f$ be the function given by $f(x) = \cos(x^2 + x) + 2$. the derivative of $f$ is given by $f(x) = -(2x + 1)\sin(x^2 + x)$. what value of $c$ satisfies the conclusion of the mean value theorem applied to $f$ on the interval $1, 2$?

option a: $1.079$ because $f(1.079) = \frac{f(2) - f(1)}{2 - 1}$

option b: $1.438$ because $f(1.438) = \frac{f(2) - f(1)}{2 - 1}$

option c: $1.750$ because $f(1.750) = 0$

option d: $1.932$ because $f(1.932) = f(1.932)$

Explanation:

Step1: Recall Mean Value Theorem (MVT)

The Mean Value Theorem states that if a function \( f(x) \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c)=\frac{f(b)-f(a)}{b - a} \). Here, \( a = 1 \), \( b = 2 \), so we need to find \( c \) in \((1, 2)\) where \( f'(c)=\frac{f(2)-f(1)}{2 - 1}=f(2)-f(1) \).

Step2: Analyze each option

• Option A: Claims \( f(1.079)=\frac{f(2)-f(1)}{2 - 1} \), but MVT involves the derivative, not the function value. Eliminate A.
• Option B: Claims \( f'(1.438)=\frac{f(2)-f(1)}{2 - 1} \), which matches the MVT conclusion (derivative at \( c \) equals the average rate of change).
• Option C: Involves the second derivative \( f''(1.750) = 0 \), which is related to concavity (Inflection point), not MVT. Eliminate C.
• Option D: Claims \( f'(1.932)=f(1.932) \), which has no relation to MVT. Eliminate D.

Answer:

B. 1.438 because \( f'(1.438)=\frac{f(2)-f(1)}{2 - 1} \)