The value of c in the Lagrange's mean value theorem for the function f (x) = x³ - 4x² + 8x + 11 when

Q: Wait, why do we need to check if the function is continuous and differentiable? Can't we just jump to the formula?

That is a fantastic question! In mathematics, theorems are like contracts—they only hold if the conditions are met. LMVT requires the function to be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$. If the function had a sharp corner or a break (discontinuity), the theorem might fail. Since our function is a polynomial, it is perfectly smooth, so the 'contract' is valid!

Q: I got two values for $c$. How do I know which one to pick?

Great job getting both roots! Remember, LMVT guarantees that $c$ exists *strictly* inside the interval $(a, b)$. In this case, our interval is $(0, 1)$. When you solve the quadratic, you will find one root is greater than 1 and one is between 0 and 1. We must reject the root outside the interval because the theorem specifically constrains $c$ to be between the endpoints.

Q: Can we use Rolle's Theorem instead of LMVT here?

Not in this case. Rolle's Theorem is actually a special case of LMVT where $f(a) = f(b)$, which would make the secant slope zero. Here, $f(0) = 11$ and $f(1) = 16$. Since $f(0) \neq f(1)$, the secant line is not horizontal, so we must use the general Lagrange's Mean Value Theorem.

Question

The value of $c$ in the Lagrange's mean value theorem for the function $f (x) = x^3 - 4x^2 + 8x + 11$ when $x \in [0, 1]$ is :

Options:

(A) $\frac{4-\sqrt{7}}{3}$

(B) $\frac{4-\sqrt{5}}{3}$

(C) $\frac{4+\sqrt{7}}{3}$

(D) $\frac{4+\sqrt{5}}{3}$

Sigma Percentile · Accepted Answer

### The Geometry of Change: Unlocking Lagrange's Mean Value Theorem

Imagine you are driving a car along a winding mountain road. Your position is defined by a function $f(x)$, and your journey takes you from point $x=0$ to $x=1$.

At the start, you are at a certain altitude, and by the end, you have climbed to a higher one. Lagrange's Mean Value Theorem (LMVT) is the mathematical bridge that connects your average speed over the entire trip to the specific moment when your speedometer reading exactly matches that average.

It is a profound realization that somewhere along that path, you were traveling at exactly your average speed.

### The Setup: A Well-Behaved Function

We are working with the cubic polynomial $f(x) = x^3 - 4x^2 + 8x + 11$ on the interval $[0, 1]$. In the world of calculus, polynomials are the most well-behaved citizens.

They are continuous and differentiable everywhere. Because our function is a polynomial, we are guaranteed that it is continuous on $[0, 1]$ and differentiable on $(0, 1)$.

This satisfies the conditions for LMVT, allowing us to proceed with confidence.

### The Secant Line: The Average Slope

To find the average rate of change, we look at the endpoints. First, we evaluate the function at the start of our interval, $x=0$:

$$f(0) = (0)^3 - 4(0)^2 + 8(0) + 11 = 11$$

Next, we evaluate it at the end, $x=1$:

$$f(1) = (1)^3 - 4(1)^2 + 8(1) + 11 = 16$$

Imagine a straight line—a secant line—connecting the points $(0, 11)$ and $(1, 16)$. The slope of this line represents the average rate of change of our function over the interval. Using the slope formula $m = \frac{f(b) - f(a)}{b - a}$, we get:

$$m = \frac{16 - 11}{1 - 0} = 5$$

This value, $5$, is the target. The theorem tells us there exists a point $c$ where the tangent to the curve is parallel to this secant line, meaning the derivative at $c$ must also be $5$.

### The Derivative: The Instantaneous Slope

Now, let us find the general expression for the slope of the tangent at any point $x$. We differentiate $f(x)$ with respect to $x$:

$$f'(x) = \frac{d}{dx}(x^3 - 4x^2 + 8x + 11) = 3x^2 - 8x + 8$$

This derivative, $f'(x) = 3x^2 - 8x + 8$, gives us the slope of the curve at any point $x$. To find the specific point $c$ where the tangent is parallel to our secant line, we set $f'(c) = 5$:

$$3c^2 - 8c + 8 = 5$$

### The Moment of Truth: Solving for $c$

Subtracting $5$ from both sides, we arrive at a standard quadratic equation:

$$3c^2 - 8c + 3 = 0$$

To solve for $c$, we use the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Substituting $a=3$, $b=-8$, and $c=3$:

$$c = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(3)}}{2(3)} = \frac{8 \pm \sqrt{64 - 36}}{6} = \frac{8 \pm \sqrt{28}}{6}$$

Simplifying $\sqrt{28}$ to $2\sqrt{7}$, we get:

$$c = \frac{8 \pm 2\sqrt{7}}{6} = \frac{4 \pm \sqrt{7}}{3}$$

### The Final Selection: Filtering the Results

We have two potential candidates for $c$: $c_1 = \frac{4 + \sqrt{7}}{3}$ and $c_2 = \frac{4 - \sqrt{7}}{3}$. Since $\sqrt{7} \approx 2.64$, we can estimate these values:

$$c_1 \approx \frac{4 + 2.64}{3} \approx 2.21$$

$$c_2 \approx \frac{4 - 2.64}{3} \approx 0.45$$

Our interval is $(0, 1)$. Clearly, $c_1$ lies outside this range, while $c_2$ sits perfectly inside.

Thus, the value of $c$ that satisfies Lagrange's Mean Value Theorem is **$c = \frac{4 - \sqrt{7}}{3}$**. You have successfully navigated the geometry of the curve to find the exact point where the instantaneous rate of change matches the average!

Animated Solution for Mathematics - Differentiation: The value of $c$ in the Lagrange's mean value theorem for the function $f (x) = x^{3} - 4 x^{2} + 8 x + 11$ when $x \in [0, 1]$ is :

Select Answer:

Visualized Solution

Function and Interval

LMVT Conditions

Evaluating Endpoints

Secant Line Slope

Differentiating $f (x)$

Applying LMVT

Forming the Quadratic

Quadratic Formula

Simplifying Roots

Selecting Valid $c$

The Sigma Insight: Mean Value Theorems

The Geometry of Change

The Setup

The Secant Line

The Derivative

The Moment of Truth

The Final Selection

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of c in the Lagrange's mean value theorem for the function f(x)=x3−4x2+8x+11 when x∈[0,1] is :