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

Q: Wait, why do we reject the root $\frac{4 + \sqrt{7}}{3}$?

Great observation! Lagrange’s Mean Value Theorem specifically guarantees that the point $c$ exists *within* the open interval $(a, b)$. In our case, the interval is $(0, 1)$. Since $\sqrt{7} \approx 2.64$, the root $\frac{4 + 2.64}{3} \approx 2.21$ lies outside our interval. We must discard it because the theorem only makes a claim about the behavior of the function *between* the endpoints.

Q: Can I use Rolle's Theorem here instead?

Rolle's Theorem is actually a special case of the Mean Value Theorem where $f(a) = f(b)$. If you check our endpoints, $f(0) = 11$ and $f(1) = 16$. Since $f(0) \neq f(1)$, Rolle's Theorem does not apply. We must use the more general Lagrange’s Mean Value Theorem.

Q: What if the function wasn't a polynomial? Would the steps change?

The steps would remain logically identical, but you would have to verify the conditions first! LMVT requires the function to be continuous on $[a, b]$ and differentiable on $(a, b)$. For polynomials, this is always true. For other functions (like those with denominators or square roots), you must ensure the function is well-behaved within the interval before proceeding.

Question

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

Options:

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

(B) $\frac{2}{3}$

(C) $\frac{\sqrt{7}-2}{3}$

(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 from point $A$ to point $B$. If you start at $x = 0$ and end at $x = 1$, and your average speed over that journey is $5$ km/h, common sense tells you that at some specific moment during that trip, your speedometer must have read exactly $5$ km/h.

This is the intuitive heart of Lagrange’s Mean Value Theorem (LMVT). It is the bridge between the average rate of change and the instantaneous rate of change.

Today, we are going to apply this beautiful concept to the cubic function $f(x) = x^3 - 4x^2 + 8x + 11$ on the interval $[0, 1]$.

### The Green Light: Checking Conditions

Before we dive into the algebra, we must respect the mathematical rigor. LMVT is not a universal law for every function; it requires a smooth, unbroken path.

Because our function $f(x)$ is a polynomial, we are blessed with continuity on the closed interval $[0, 1]$ and differentiability on the open interval $(0, 1)$. The path is smooth, the conditions are met, and we have the green light to proceed.

### The Secant Line: Defining the Average

We begin by anchoring our journey. We need to know where we start and where we finish.

At $x = 0$, our function value is $f(0) = 11$. At $x = 1$, our function value is $f(1) = 16$.

These two points, $(0, 11)$ and $(1, 16)$, define the secant line—the straight path connecting our start and end. The slope of this line represents the average rate of change:

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

This value, $5$, is the target slope we are looking for on the curve.

### The Tangent Line: Finding the Instantaneous

Now, we need to find where the curve mimics this average slope. We need the derivative, $f'(x)$, which tells us the slope of the tangent at any point $x$.

Using the power rule, we differentiate $f(x) = x^3 - 4x^2 + 8x + 11$ to get:

$$f'(x) = 3x^2 - 8x + 8$$

This expression is our tool to find the instantaneous slope. According to LMVT, there exists a point $c$ in $(0, 1)$ such that the slope of the tangent at $c$ is equal to the slope of the secant line. So, we set $f'(c) = 5$.

### The Quadratic Challenge

This leads us to the equation $3c^2 - 8c + 8 = 5$. Subtracting $5$ from both sides, we arrive at the quadratic equation:

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

This is the moment where many students panic, but stay calm. We use the quadratic formula:

$$c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Substituting our coefficients $a = 3$, $b = -8$, and $c_{const} = 3$, we get:

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

Simplifying the discriminant, we have $64 - 36 = 28$. Thus:

$$c = \frac{8 \pm \sqrt{28}}{6} = \frac{8 \pm 2\sqrt{7}}{6} = \frac{4 \pm \sqrt{7}}{3}$$

### The Final Selection

We have two potential candidates for $c$: $c_1 = \frac{4 + \sqrt{7}}{3}$ and $c_2 = \frac{4 - \sqrt{7}}{3}$. We must return to our constraint: $c \in (0, 1)$.

If we approximate $\sqrt{7} \approx 2.64$, then $c_1 \approx \frac{4 + 2.64}{3} \approx 2.21$, which is clearly outside our interval. However, $c_2 \approx \frac{4 - 2.64}{3} \approx 0.45$, which fits perfectly within $(0, 1)$.

We have found our point! The value of $c$ that satisfies the theorem is:

$$\mathbf{c = \frac{4 - \sqrt{7}}{3}}$$

You have successfully navigated the geometry, the calculus, and the algebraic constraints. This is the power of mathematics—turning an abstract theorem into a precise, calculated reality.

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

Select Answer:

Visualized Solution

The Function $f (x)$

LMVT Conditions

Evaluating Endpoints

Slope of the Secant Line

Finding the Derivative $f^{'} (x)$

Applying LMVT

Setting up the Equation

Solving for $c$

Simplifying the Roots

Selecting the Valid Interval

The Sigma Insight: Mean Value Theorems

The Geometry of Change

The Green Light

The Secant Line

The Tangent Line

The Quadratic Challenge

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 Lagrange’s mean value theorem for the function f(x)=x3−4x2+8x+11, where x∈[0,1] is :