Math Problem Statement

Show that the function f: [1,4]→R defined by f(x) = x ^ 3 - 3x is an increasing function

Solution

To show that the function f(x)=x33xf(x) = x^3 - 3x is increasing on the interval [1,4][1,4], we need to examine the behavior of its derivative. A function is increasing on an interval if its derivative is non-negative throughout that interval.

Step 1: Compute the derivative of f(x)f(x)

The derivative of f(x)=x33xf(x) = x^3 - 3x is:

f(x)=ddx(x33x)=3x23f'(x) = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3

Step 2: Analyze the sign of f(x)f'(x)

We need to determine where f(x)=3x23f'(x) = 3x^2 - 3 is positive, negative, or zero.

First, let's factor the derivative:

f(x)=3(x21)=3(x1)(x+1)f'(x) = 3(x^2 - 1) = 3(x - 1)(x + 1)

Now, let's check the critical points by setting f(x)=0f'(x) = 0:

3(x1)(x+1)=03(x - 1)(x + 1) = 0

This equation gives two critical points:

x=1andx=1x = 1 \quad \text{and} \quad x = -1

However, since the function is defined on the interval [1,4][1,4], we are only concerned with x=1x = 1 in this interval.

Step 3: Determine the sign of f(x)f'(x) on the interval [1,4][1,4]

Now, let's examine the sign of f(x)=3(x1)(x+1)f'(x) = 3(x - 1)(x + 1) on the interval [1,4][1, 4]:

  • At x=1x = 1, f(x)=3(11)(1+1)=0f'(x) = 3(1 - 1)(1 + 1) = 0.
  • For x>1x > 1 (in particular, for x(1,4]x \in (1, 4]), both (x1)>0(x - 1) > 0 and (x+1)>0(x + 1) > 0, so f(x)>0f'(x) > 0.

Thus, f(x)0f'(x) \geq 0 for all x[1,4]x \in [1, 4], with equality only at x=1x = 1.

Conclusion:

Since f(x)0f'(x) \geq 0 on [1,4][1,4], the function f(x)=x33xf(x) = x^3 - 3x is increasing on the interval [1,4][1,4].

Would you like further clarification on any step or more details?

Here are some related questions for further exploration:

  1. How do critical points influence the behavior of a function?
  2. What does it mean for a function to be strictly increasing versus non-decreasing?
  3. How can the second derivative provide more insights into the nature of a function's increase or decrease?
  4. What is the role of endpoints when determining if a function is increasing or decreasing on a closed interval?
  5. Can a function be increasing on an interval but have a flat (zero derivative) point within that interval?

Tip: A derivative being zero at a point does not mean the function decreases or increases—it just indicates a potential critical point (like a local maximum, minimum, or inflection point). Always check the sign of the derivative around critical points!

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Increasing Functions

Formulas

f'(x) = 3x^2 - 3

Theorems

First Derivative Test

Suitable Grade Level

Grades 11-12 or Early College