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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
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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
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