168k views
5 votes
Find the exact location of all the relative and absolute extrema of the function f(x) = x² - 4x + 1 with domain [0, 3].

User Mfluehr
by
8.4k points

1 Answer

6 votes

Final answer:

To locate the extrema of the function f(x) = x² - 4x + 1, you find the derivative and solve for critical points, yielding x=2 as the sole critical point. Evaluating f(x) at x=0, x=2, and x=3 within the domain [0, 3] indicates f(2)=-3 as the minimum and f(3)=2 as a relative maximum.

Step-by-step explanation:

To find the relative and absolute extrema of the function f(x) = x² - 4x + 1 with domain [0, 3], we start by taking the derivative of f(x) to locate the critical points. The derivative f'(x) = 2x - 4. Setting this equal to zero, we solve for x to find the critical points: 2x - 4 = 0, resulting in x = 2.

Next, we evaluate the function at the critical point and the endpoints of the domain to determine the extrema:

  1. f(0) = 0² - 4· 0 + 1 = 1
  2. f(2) = 2² - 4· 2 + 1 = -3
  3. f(3) = 3² - 4· 3 + 1 = 2

From these values, f(2) = -3 is the minimum (both relative and absolute), since it is the lowest value of the function on our closed interval [0, 3], and f(3) = 2 is a relative maximum, but not the absolute maximum since f(0) = 1 is the highest value on the interval.

User Giacomo Catenazzi
by
8.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories