Math Problem Statement

Consider a function f(x)=a x^{3}+b x^{2}+c x+d, with three real roots x_{0}, x_{1}, x_{2} such that x_{0}<x_{1}<x_{2}. Show that there must be a local maximum between x_{0} and x_{1} and a local minimum between x_{1} and x_{2}.

Solution

Given the function f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d with real roots x0x_0, x1x_1, and x2x_2 such that x0<x1<x2x_0 < x_1 < x_2, we are asked to show that there must be a local maximum between x0x_0 and x1x_1 and a local minimum between x1x_1 and x2x_2.

Step 1: Analyze the first derivative of f(x)f(x)

The first step is to find the critical points of f(x)f(x) by taking its first derivative:

f(x)=ddx(ax3+bx2+cx+d)=3ax2+2bx+cf'(x) = \frac{d}{dx}(ax^3 + bx^2 + cx + d) = 3ax^2 + 2bx + c

The critical points occur where f(x)=0f'(x) = 0, i.e.,

3ax2+2bx+c=03ax^2 + 2bx + c = 0

This is a quadratic equation, and its roots (let's call them r1r_1 and r2r_2) can be found using the quadratic formula:

r1,r2=2b±(2b)24(3a)(c)2(3a)=2b±4b212ac6a=b±b23ac3ar_1, r_2 = \frac{-2b \pm \sqrt{(2b)^2 - 4(3a)(c)}}{2(3a)} = \frac{-2b \pm \sqrt{4b^2 - 12ac}}{6a} = \frac{-b \pm \sqrt{b^2 - 3ac}}{3a}

Step 2: Understand the significance of the critical points r1r_1 and r2r_2

The roots r1r_1 and r2r_2 are the critical points where f(x)f(x) could have local extrema (maximum or minimum). Since f(x)f(x) is a cubic polynomial, it will have at most two real critical points.

Step 3: Location of the critical points relative to the roots of f(x)f(x)

Given that the function f(x)f(x) has three distinct real roots x0x_0, x1x_1, and x2x_2 (with x0<x1<x2x_0 < x_1 < x_2), the following must hold:

  • Between x0x_0 and x1x_1, f(x)f(x) changes from positive to negative (since f(x0)=0f(x_0) = 0 and f(x1)=0f(x_1) = 0).
  • Between x1x_1 and x2x_2, f(x)f(x) changes from negative to positive (since f(x1)=0f(x_1) = 0 and f(x2)=0f(x_2) = 0).

For f(x)f(x) to have such changes in sign between the roots, f(x)f'(x) must change sign, which implies that there is a critical point in each of these intervals:

  1. There must be a critical point r1r_1 between x0x_0 and x1x_1, where f(x)=0f'(x) = 0 and f(x)f'(x) changes from positive to negative, indicating a local maximum.

  2. There must be a critical point r2r_2 between x1x_1 and x2x_2, where f(x)=0f'(x) = 0 and f(x)f'(x) changes from negative to positive, indicating a local minimum.

Step 4: Conclusion

Therefore, there must be a local maximum between x0x_0 and x1x_1, and a local minimum between x1x_1 and x2x_2. This is a consequence of the Intermediate Value Theorem, which ensures that the continuous derivative f(x)f'(x) must cross zero in these intervals, combined with the fact that f(x)f(x) changes sign between the roots.

Would you like further details or have any questions?

Here are some related questions you might find useful:

  1. What is the significance of the discriminant in determining the nature of the roots for cubic functions?
  2. How do the roots of the first derivative relate to the concavity of the original function?
  3. Can a cubic function have more than one local maximum or minimum?
  4. How can you determine the inflection point of a cubic function?
  5. What happens to the nature of the critical points if the roots of the cubic function are not distinct?

Tip: When analyzing cubic functions, always check the first derivative to find critical points and the second derivative to determine the nature (max/min) of these points.

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Math Problem Analysis

Mathematical Concepts

Polynomials
Cubic Functions
Derivatives
Local Extrema

Formulas

Quadratic formula

Theorems

Intermediate Value Theorem

Suitable Grade Level

Grades 11-12