Math Problem Statement

Verify Lagrange mean value for f(x) = x ^ 3 + x ^ 2 in [1,2]

Solution

To verify the Lagrange Mean Value Theorem (LMVT) for the function f(x)=x3+x2f(x) = x^3 + x^2 on the interval [1,2][1, 2], let's proceed step by step.

Step 1: Verify Conditions of LMVT

LMVT states that if a function f(x)f(x) is:

  1. Continuous on a closed interval [a,b][a, b],
  2. Differentiable on an open interval (a,b)(a, b),

then there exists at least one point c(a,b)c \in (a, b) such that: f(c)=f(b)f(a)baf'(c) = \frac{f(b) - f(a)}{b - a}

The function f(x)=x3+x2f(x) = x^3 + x^2 is a polynomial, which is continuous and differentiable everywhere, so it satisfies both conditions on [1,2][1, 2].

Step 2: Calculate f(1)f(1) and f(2)f(2)

Let's find f(1)f(1) and f(2)f(2) to compute the slope of the secant line between these points: f(1)=13+12=1+1=2f(1) = 1^3 + 1^2 = 1 + 1 = 2 f(2)=23+22=8+4=12f(2) = 2^3 + 2^2 = 8 + 4 = 12

Step 3: Calculate the Secant Slope

The secant slope between x=1x = 1 and x=2x = 2 is: f(2)f(1)21=12221=101=10\frac{f(2) - f(1)}{2 - 1} = \frac{12 - 2}{2 - 1} = \frac{10}{1} = 10

Step 4: Find f(x)f'(x)

Now, we calculate the derivative f(x)f'(x): f(x)=x3+x2f(x) = x^3 + x^2 f(x)=3x2+2xf'(x) = 3x^2 + 2x

Step 5: Set f(c)=10f'(c) = 10 and Solve for cc

We now need to find c(1,2)c \in (1, 2) such that f(c)=10f'(c) = 10: 3c2+2c=103c^2 + 2c = 10 Rearrange to form a standard quadratic equation: 3c2+2c10=03c^2 + 2c - 10 = 0

Solve this quadratic equation using the quadratic formula: c=2±(2)243(10)23c = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 3 \cdot (-10)}}{2 \cdot 3} c=2±4+1206c = \frac{-2 \pm \sqrt{4 + 120}}{6} c=2±1246c = \frac{-2 \pm \sqrt{124}}{6} c=2±2316c = \frac{-2 \pm 2\sqrt{31}}{6} c=1±313c = \frac{-1 \pm \sqrt{31}}{3}

Calculating these values: c \approx 1.62 \quad \text{(since 1 < c < 2)}

Conclusion

There exists a point c1.62c \approx 1.62 in the interval (1,2)(1, 2) where f(c)=10f'(c) = 10, thus satisfying the Lagrange Mean Value Theorem.

Would you like more details or have any questions?

Here are some related questions:

  1. How does the Mean Value Theorem differ from Rolle's Theorem?
  2. Can LMVT be applied to trigonometric functions over specified intervals?
  3. What are the implications of LMVT in real-world scenarios?
  4. How does LMVT apply to non-polynomial functions?
  5. How can we approximate values using LMVT in calculus problems?

Tip: Always check the continuity and differentiability of the function over the interval to verify conditions for applying LMVT.

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Math Problem Analysis

Mathematical Concepts

Calculus
Mean Value Theorem
Derivatives
Polynomials

Formulas

Lagrange Mean Value Theorem formula: f'(c) = (f(b) - f(a)) / (b - a)
Derivative of f(x) = x^3 + x^2: f'(x) = 3x^2 + 2x

Theorems

Lagrange Mean Value Theorem

Suitable Grade Level

Grades 11-12