Question

Consider a function $f(x) = x^2 + 2x + 3$ on the interval $[-2,1]$. The value of $c$ that satisfies the conclusion of Lagrange's Mean Value Theorem is:

• A. -1/2
• B. 0
• C. 1/2
• D. 1

Answer 1

Answer: (A)

-1/2

Answer 2

Lagrange's Mean Value Theorem (LMVT):

If a function $f(x)$ is continuous on the 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:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

Given function: $f(x) = x^2 + 2x + 3$

Given interval: $[-2, 1]$, so $a = -2$ and $b = 1$.

Step 1: Check the conditions for LMVT.

Since $f(x)$ is a polynomial function, it is continuous on $[-2, 1]$ and differentiable on $(-2, 1)$. Thus, the conditions for LMVT are satisfied.

Step 2: Calculate $f(a)$ and $f(b)$.

$f(a) = f(-2) = (-2)^2 + 2(-2) + 3 = 4 - 4 + 3 = 3$

$f(b) = f(1) = (1)^2 + 2(1) + 3 = 1 + 2 + 3 = 6$

Step 3: Calculate the slope of the secant line.

$\frac{f(b) - f(a)}{b - a} = \frac{6 - 3}{1 - (-2)} = \frac{3}{1 + 2} = \frac{3}{3} = 1$

Step 4: Calculate the derivative $f'(x)$.

$f'(x) = \frac{d}{dx}(x^2 + 2x + 3) = 2x + 2$

Step 5: Set $f'(c)$ equal to the slope of the secant line and solve for $c$.

$f'(c) = 2c + 2$

According to LMVT:

$2c + 2 = 1$

$2c = 1 - 2$

$2c = -1$

$c = -\frac{1}{2}$

Step 6: Verify if $c$ lies within the open interval $(a, b)$.

The open interval is $(-2, 1)$.

Since $c = -\frac{1}{2} = -0.5$, and $-2 < -0.5 < 1$, the value of $c$ is indeed within the interval.

The value of $c$ that satisfies the conclusion of Lagrange's Mean Value Theorem is $-\frac{1}{2}$.