Math Problem Statement

Let f(x)=x3−6x2+3f(x)=x3−6x2+3

Find the critical numbers of ff: 

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 (Separate multiple answers by commas.)

Determine the intervals on which ff is increasing and decreasing. Help entering intervalsff is increasing on: 

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ff is decreasing on: 

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Use the First Derivative Test to determine whether each critical point is a relative maximum, minimum, or neither. (Separate multiple answers by commas, if there is no answer enter “none”.)Relative maxima occur at x=x=

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Relative minima occur at x=x=

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Solution

To analyze the function f(x)=x36x2+3f(x) = x^3 - 6x^2 + 3, we need to find its critical numbers, determine the intervals on which it is increasing or decreasing, and use the First Derivative Test to classify the critical points.

Step 1: Finding Critical Numbers

Critical numbers occur where the derivative f(x)f'(x) is zero or undefined.

  1. Compute the first derivative f(x)f'(x): f(x)=ddx(x36x2+3)=3x212xf'(x) = \frac{d}{dx}(x^3 - 6x^2 + 3) = 3x^2 - 12x
  2. Set the derivative equal to zero and solve for xx: 3x212x=03x^2 - 12x = 0 Factor out 3x3x: 3x(x4)=03x(x - 4) = 0 So the critical numbers are: x=0,x=4x = 0, \quad x = 4

Step 2: Determining Intervals of Increase and Decrease

To determine where f(x)f(x) is increasing or decreasing, we analyze the sign of f(x)f'(x) on the intervals determined by the critical points.

  1. Choose test points in the intervals (,0)(-\infty, 0), (0,4)(0, 4), and (4,)(4, \infty).

    • For x(,0)x \in (-\infty, 0): f(1)=3(1)212(1)=3+12=15(positive)f'(-1) = 3(-1)^2 - 12(-1) = 3 + 12 = 15 \quad (\text{positive})
    • For x(0,4)x \in (0, 4): f(2)=3(2)212(2)=1224=12(negative)f'(2) = 3(2)^2 - 12(2) = 12 - 24 = -12 \quad (\text{negative})
    • For x(4,)x \in (4, \infty): f(5)=3(5)212(5)=7560=15(positive)f'(5) = 3(5)^2 - 12(5) = 75 - 60 = 15 \quad (\text{positive})

Therefore, f(x)f(x) is:

  • Increasing on (,0)(4,)(-\infty, 0) \cup (4, \infty)
  • Decreasing on (0,4)(0, 4)

Step 3: First Derivative Test for Relative Extrema

Use the First Derivative Test to classify the critical points x=0x = 0 and x=4x = 4.

  1. At x=0x = 0:

    • f(x)f'(x) changes from positive to negative.
    • Hence, x=0x = 0 is a relative maximum.
  2. At x=4x = 4:

    • f(x)f'(x) changes from negative to positive.
    • Hence, x=4x = 4 is a relative minimum.

Summary

  • Critical numbers: x=0,4x = 0, 4
  • Increasing intervals: (,0)(4,)(-\infty, 0) \cup (4, \infty)
  • Decreasing intervals: (0,4)(0, 4)
  • Relative maxima: x=0x = 0
  • Relative minima: x=4x = 4

Would you like more details or have any questions?

Here are some related questions:

  1. How do we find the second derivative of f(x)f(x) and use it to confirm concavity?
  2. What is the behavior of f(x)f(x) as xx approaches ±\pm\infty?
  3. Can we identify any points of inflection for f(x)f(x)?
  4. How do the relative extrema of f(x)f(x) relate to the overall graph of the function?
  5. What are the coordinates of the relative extrema points for f(x)f(x)?
  6. How can we sketch the graph of f(x)f(x) using the critical points and intervals of increase/decrease?
  7. What is the absolute maximum and minimum of f(x)f(x) on a closed interval, e.g., [0,5][0, 5]?
  8. How does the graph of f(x)=x36x2+3f(x) = x^3 - 6x^2 + 3 compare to other cubic functions?

Tip: When analyzing a function, always check for critical points and intervals of increase/decrease to understand its behavior thoroughly.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Critical Points
First Derivative Test

Formulas

First derivative formula: f'(x) = 3x^2 - 12x

Theorems

First Derivative Test

Suitable Grade Level

Advanced High School