Math Problem Statement

Consider the function f(x) = 4 - 6x^2 on the interval -5 ≤ x ≤ 1. The absolute maximum value is ___ and this occurs at x = ___. The absolute minimum value is ___ and this occurs at x = ___.

Solution

We are given the function f(x)=46x2f(x) = 4 - 6x^2 on the interval 5x1-5 \leq x \leq 1. To find the absolute maximum and minimum values, we'll follow these steps:

1. Find the critical points:

We first take the derivative of f(x)f(x) to find where the slope is zero (critical points).

f(x)=ddx(46x2)=12xf'(x) = \frac{d}{dx} (4 - 6x^2) = -12x

Set the derivative equal to zero to find critical points:

12x=0    x=0-12x = 0 \implies x = 0

So, x=0x = 0 is a critical point.

2. Evaluate the function at the endpoints and critical points:

The function must also be evaluated at the boundaries of the interval, x=5x = -5 and x=1x = 1, and at the critical point x=0x = 0.

  • At x=5x = -5: f(5)=46(5)2=46(25)=4150=146f(-5) = 4 - 6(-5)^2 = 4 - 6(25) = 4 - 150 = -146

  • At x=0x = 0: f(0)=46(0)2=4f(0) = 4 - 6(0)^2 = 4

  • At x=1x = 1: f(1)=46(1)2=46=2f(1) = 4 - 6(1)^2 = 4 - 6 = -2

3. Determine the maximum and minimum values:

Now, we compare the values of f(x)f(x) at the critical point and endpoints:

  • f(5)=146f(-5) = -146
  • f(0)=4f(0) = 4
  • f(1)=2f(1) = -2

The absolute maximum value is f(0)=4f(0) = 4, and this occurs at x=0x = 0.

The absolute minimum value is f(5)=146f(-5) = -146, and this occurs at x=5x = -5.

Final Answers:

  • Absolute maximum value: 44 at x=0x = 0.
  • Absolute minimum value: 146-146 at x=5x = -5.

Would you like more details on any step or have any questions?


Here are 5 questions related to this topic:

  1. How does the second derivative help in classifying critical points?
  2. What is the geometric interpretation of finding maximum and minimum values on a given interval?
  3. Can a function have more than one absolute maximum or minimum value?
  4. How do constraints on an interval affect the evaluation of extrema?
  5. Why is it important to check both critical points and endpoints in a closed interval?

Tip: Always check both the critical points and the endpoints when finding absolute extrema on a closed interval.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Optimization
Derivatives
Critical Points

Formulas

f(x) = 4 - 6x^2
f'(x) = -12x
f'(x) = 0 for critical points

Theorems

First Derivative Test
Extreme Value Theorem

Suitable Grade Level

Grades 11-12