Math Problem Statement

Consider the polynomial function f(x) = x^5 - 3x^3 + 4x. (a) Find the intervals on which f is increasing. (b) Find the intervals on which f is decreasing.

Solution

To solve the given problem, we need to analyze the intervals of increase and decrease for the polynomial function f(x)=x53x3+4xf(x) = x^5 - 3x^3 + 4x.

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

The first derivative, f(x)f'(x), helps identify critical points and the behavior of f(x)f(x): f(x)=5x49x2+4f'(x) = 5x^4 - 9x^2 + 4

Step 2: Solve f(x)=0f'(x) = 0 to find critical points

Set f(x)=0f'(x) = 0: 5x49x2+4=05x^4 - 9x^2 + 4 = 0 Let u=x2u = x^2, so the equation becomes: 5u29u+4=05u^2 - 9u + 4 = 0 Solve using the quadratic formula: u=b±b24ac2a, where a=5,b=9,c=4u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \text{ where } a = 5, b = -9, c = 4 u=9±(9)24(5)(4)2(5)u = \frac{9 \pm \sqrt{(-9)^2 - 4(5)(4)}}{2(5)} u=9±818010u = \frac{9 \pm \sqrt{81 - 80}}{10} u=9±110u = \frac{9 \pm 1}{10} u=1oru=0.8u = 1 \quad \text{or} \quad u = 0.8 Thus, x2=1x^2 = 1 or x2=0.8x^2 = 0.8. This gives: x=±1,x=±0.8x = \pm 1, \quad x = \pm \sqrt{0.8}

Step 3: Determine the sign of f(x)f'(x) in each interval

The critical points divide the real number line into intervals: (,1)(-\infty, -1), (1,0.8)(-1, -\sqrt{0.8}), (0.8,0.8)(-\sqrt{0.8}, \sqrt{0.8}), (0.8,1)(\sqrt{0.8}, 1), (1,)(1, \infty). Test f(x)f'(x) in each interval:

  1. Interval (,1)(-\infty, -1): Pick x=2x = -2. Substitute into f(x)f'(x): f(2)=5(2)49(2)2+4=8036+4=48>0f'(-2) = 5(-2)^4 - 9(-2)^2 + 4 = 80 - 36 + 4 = 48 > 0 f(x)f(x) is increasing.

  2. Interval (1,0.8)(-1, -\sqrt{0.8}): Pick x=0.9x = -0.9. Substitute into f(x)f'(x): f(0.9)=5(0.9)49(0.9)2+4f'(-0.9) = 5(-0.9)^4 - 9(-0.9)^2 + 4 Calculation shows f(0.9)<0f'(-0.9) < 0. f(x)f(x) is decreasing.

  3. Interval (0.8,0.8)(-\sqrt{0.8}, \sqrt{0.8}): Pick x=0x = 0. Substitute: f(0)=5(0)49(0)2+4=4>0f'(0) = 5(0)^4 - 9(0)^2 + 4 = 4 > 0 f(x)f(x) is increasing.

  4. Interval (0.8,1)(\sqrt{0.8}, 1): Pick x=0.9x = 0.9. Substitute: Calculation shows f(0.9)<0f'(0.9) < 0. f(x)f(x) is decreasing.

  5. Interval (1,)(1, \infty): Pick x=2x = 2. Substitute: f(2)=5(2)49(2)2+4=8036+4=48>0f'(2) = 5(2)^4 - 9(2)^2 + 4 = 80 - 36 + 4 = 48 > 0 f(x)f(x) is increasing.

Step 4: Write the intervals

  • f(x)f(x) is increasing on (,1)(0.8,0.8)(1,)(-\infty, -1) \cup (-\sqrt{0.8}, \sqrt{0.8}) \cup (1, \infty).
  • f(x)f(x) is decreasing on (1,0.8)(0.8,1)(-1, -\sqrt{0.8}) \cup (\sqrt{0.8}, 1).

Would you like a step-by-step walkthrough or further clarification?


Related questions:

  1. How do we find higher-order derivatives to classify critical points further?
  2. What role does the second derivative play in determining concavity?
  3. Can you use a graphing utility to verify these intervals visually?
  4. How can you confirm the nature of turning points using f(x)f''(x)?
  5. What happens to f(x)f(x) as x±x \to \pm\infty?

Tip: Always test critical points carefully to avoid algebraic errors when solving for intervals.

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Math Problem Analysis

Mathematical Concepts

Calculus
Critical Points
Increasing and Decreasing Functions

Formulas

First derivative f'(x)
Quadratic equation formula ax^2 + bx + c = 0

Theorems

Critical point theorem
Sign test for derivative

Suitable Grade Level

Undergraduate - Calculus I