Math Problem Statement

f(x)=x^5+5x^4-35x^3 on [-1,4]. if f is increasing on the interval [-1,4], which of the following must be true? A. f(x)>0 for all x in (-1,4). B. f(x) is constant on (-1,4). C. f'(x)>0 for all x in (-1,4). D. both f'(x)>0 and f(x)>0 for all x in (-1,4)?

Solution

To determine the conditions under which f(x)=x5+5x435x3f(x) = x^5 + 5x^4 - 35x^3 is increasing on the interval [1,4][-1, 4], we need to analyze its derivative f(x)f'(x).

Step 1: Find the derivative f(x)f'(x)

The derivative is calculated as follows:

f(x)=5x4+20x3105x2f'(x) = 5x^4 + 20x^3 - 105x^2

Step 2: Analyze f(x)f'(x)

For f(x)f(x) to be increasing on the interval, we require f(x)0f'(x) \geq 0.

Step 3: Set f(x)=0f'(x) = 0

We need to find the critical points by solving:

5x4+20x3105x2=05x^4 + 20x^3 - 105x^2 = 0

Factoring out common terms:

5x2(x2+4x21)=05x^2(x^2 + 4x - 21) = 0

This gives us:

  1. x2=0x^2 = 0 (i.e., x=0x = 0)
  2. For x2+4x21=0x^2 + 4x - 21 = 0, we can use the quadratic formula:

x=4±4241(21)21=4±1002=4±102x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-21)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2}

This yields:

x=3andx=7x = 3 \quad \text{and} \quad x = -7

The relevant critical points in the interval [1,4][-1, 4] are x=0x = 0 and x=3x = 3.

Step 4: Test intervals

We will test the sign of f(x)f'(x) in the intervals [1,0][-1, 0], (0,3)(0, 3), and (3,4)(3, 4):

  1. For x[1,0]x \in [-1, 0]:

    • Choose x=1x = -1: f(1)=5(1)4+20(1)3105(1)2=520105=120<0f'(-1) = 5(-1)^4 + 20(-1)^3 - 105(-1)^2 = 5 - 20 - 105 = -120 < 0
  2. For x(0,3)x \in (0, 3):

    • Choose x=1x = 1: f(1)=5(1)4+20(1)3105(1)2=5+20105=80<0f'(1) = 5(1)^4 + 20(1)^3 - 105(1)^2 = 5 + 20 - 105 = -80 < 0
  3. For x(3,4)x \in (3, 4):

    • Choose x=4x = 4: f(4)=5(4)4+20(4)3105(4)2=5(256)+20(64)105(16)=1280+12801680=880>0f'(4) = 5(4)^4 + 20(4)^3 - 105(4)^2 = 5(256) + 20(64) - 105(16) = 1280 + 1280 - 1680 = 880 > 0

Step 5: Conclusions

  • f(x)f(x) is not increasing on the entire interval [1,4][-1, 4] because f(x)<0f'(x) < 0 in parts of that interval.
  • Therefore, none of the statements can be concluded as universally true based on ff being increasing.

Summary of Options:

  • A. False: f(x)f(x) does not have to be positive.
  • B. False: f(x)f(x) is not constant since it has varying behavior.
  • C. False: f(x)>0f'(x) > 0 is not true throughout [1,4][-1, 4].
  • D. False: Neither f(x)>0f'(x) > 0 nor f(x)>0f(x) > 0 holds in the interval.

Final Answer

None of the options (A, B, C, or D) must be true.

Would you like further details or have any questions about this analysis? Here are some follow-up questions:

  1. What is the significance of critical points in determining the behavior of a function?
  2. How can you determine the intervals where a function is increasing or decreasing?
  3. What role does the second derivative play in analyzing the behavior of a function?
  4. Can you provide examples of functions that are increasing over a specified interval?
  5. How would you graph f(x)f(x) to visually assess its increasing/decreasing behavior?

Tip: Always consider testing values around critical points to understand the behavior of a function on an interval.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Function Behavior

Formulas

f'(x) = 5x^4 + 20x^3 - 105x^2

Theorems

First Derivative Test

Suitable Grade Level

Grades 11-12