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 an increasing function on the interval [1,4][1, 4], we need to examine its derivative, as the derivative of a function tells us about its rate of change.

Step 1: Find 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)

Now, we need to determine whether f(x)0f'(x) \geq 0 for all x[1,4]x \in [1, 4]. This will tell us if f(x)f(x) is increasing throughout the interval.

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

Now, let's analyze 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(1)=3(11)(1+1)=0f'(1) = 3(1 - 1)(1 + 1) = 0.
  • For x>1x > 1, since x1>0x - 1 > 0 and x+1>0x + 1 > 0, f(x)>0f'(x) > 0.

Thus, for x[1,4]x \in [1, 4], f(x)0f'(x) \geq 0, meaning the function is non-decreasing, and since f(x)>0f'(x) > 0 for x>1x > 1, it is strictly increasing on (1,4](1, 4].

Conclusion:

The function f(x)=x33xf(x) = x^3 - 3x is increasing on the interval [1,4][1, 4] because its derivative is non-negative and positive for x>1x > 1.

Would you like more details or have any questions?

Here are 5 related questions you might want to explore:

  1. What is the difference between a strictly increasing and a non-decreasing function?
  2. How does the second derivative of a function help in determining concavity?
  3. Can a function be increasing even if its derivative is zero at some points?
  4. How do we classify critical points in a function like f(x)f(x)?
  5. What is the importance of the Mean Value Theorem in understanding function behavior?

Tip: Always check the derivative of a function over a specific interval to determine whether the function is increasing, decreasing, or constant.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Increasing and Decreasing Functions

Formulas

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

Theorems

First Derivative Test

Suitable Grade Level

Grades 11-12