Math Problem Statement

Find the exact location of all the relative and absolute extrema of the function. (Order your answers from smallest to largest x.) f(x) = 3x2 − 12x + 3 with domain [0, 3] f has at (x, y) =

. f has at (x, y) =

. f has at (x, y) =

.

Solution

To find the exact location of the relative and absolute extrema of the function f(x)=3x212x+3f(x) = 3x^2 - 12x + 3 on the interval [0,3][0, 3], we can follow these steps:

Step 1: Find the first derivative of the function

The first derivative, f(x)f'(x), will help us identify critical points where the slope of the function is zero or undefined.

f(x)=ddx(3x212x+3)=6x12f'(x) = \frac{d}{dx} (3x^2 - 12x + 3) = 6x - 12

Step 2: Set the first derivative equal to zero

To find critical points, set f(x)=0f'(x) = 0:

6x12=06x - 12 = 0

Solve for xx:

x=2x = 2

Thus, x=2x = 2 is a critical point.

Step 3: Evaluate the function at the endpoints of the domain and at the critical point

The function is defined on the closed interval [0,3][0, 3], so we need to evaluate f(x)f(x) at both endpoints of the interval as well as at the critical point x=2x = 2.

  • f(0)=3(0)212(0)+3=3f(0) = 3(0)^2 - 12(0) + 3 = 3
  • f(2)=3(2)212(2)+3=1224+3=9f(2) = 3(2)^2 - 12(2) + 3 = 12 - 24 + 3 = -9
  • f(3)=3(3)212(3)+3=2736+3=6f(3) = 3(3)^2 - 12(3) + 3 = 27 - 36 + 3 = -6

Step 4: Determine the relative and absolute extrema

Now, we compare the function values at the critical point and the endpoints:

  • f(0)=3f(0) = 3

  • f(2)=9f(2) = -9

  • f(3)=6f(3) = -6

  • The absolute maximum value is 33, which occurs at x=0x = 0.

  • The absolute minimum value is 9-9, which occurs at x=2x = 2.

  • The relative maximum is not present because the function is continuously decreasing after the critical point x=2x = 2.

Final Answer

  1. f(x)f(x) has an absolute maximum at (0,3)(0, 3).
  2. f(x)f(x) has a relative minimum at (2,9)(2, -9).
  3. f(x)f(x) has an absolute minimum at (2,9)(2, -9).

Would you like more details or have any further questions on this? Here are some related topics to explore:

  1. How to identify critical points of a function.
  2. The first and second derivative tests for extrema.
  3. How to check for concavity and inflection points.
  4. Understanding the difference between relative and absolute extrema.
  5. The concept of a function's behavior on a closed interval.

Tip: Always check the endpoints of a function on a closed interval because the absolute extrema could occur there, even if there are critical points inside the interval.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Extrema
Critical Points
Function Behavior on a Closed Interval

Formulas

f'(x) = 6x - 12
f(x) = 3x^2 - 12x + 3

Theorems

First Derivative Test
Closed Interval Test

Suitable Grade Level

Grades 11-12