Math Problem Statement

Diberikan fungsi y = x^3 - 3x^2 + 3. Tentukan: a) Untuk interval mana y naik dan turun, b) Titik maksimum, titik minimum, dan titik belok dari y, c) Gambar sketsa grafik fungsi y.

Solution

This is a calculus problem involving the function y=x33x2+3y = x^3 - 3x^2 + 3. Let us solve it step by step.


Problem Breakdown:

  1. Given function: y=x33x2+3y = x^3 - 3x^2 + 3.

a) Determine intervals where yy is increasing or decreasing

To find intervals of increase and decrease, calculate the first derivative and analyze its sign: y=dydx=3x26x.y' = \frac{dy}{dx} = 3x^2 - 6x.

Factorize: y=3x(x2).y' = 3x(x - 2).

Critical points (where y=0y' = 0): x=0andx=2.x = 0 \quad \text{and} \quad x = 2.

Analyze the sign of yy' in the intervals:

  • (,0)(-\infty, 0): Choose x=1x = -1, y=3(1)(12)=9>0y' = 3(-1)(-1 - 2) = 9 > 0 → Increasing.
  • (0,2)(0, 2): Choose x=1x = 1, y=3(1)(12)=3<0y' = 3(1)(1 - 2) = -3 < 0 → Decreasing.
  • (2,)(2, \infty): Choose x=3x = 3, y=3(3)(32)=9>0y' = 3(3)(3 - 2) = 9 > 0 → Increasing.

Thus, yy is:

  • Increasing on (,0)(2,)(-\infty, 0) \cup (2, \infty).
  • Decreasing on (0,2)(0, 2).

b) Find maximum, minimum points, and inflection points

  1. Critical Points: Substitute x=0x = 0 and x=2x = 2 into the original function y(x)y(x) to find the corresponding yy-values:

    • At x=0x = 0: y=033(02)+3=3y = 0^3 - 3(0^2) + 3 = 3. This is a local maximum.
    • At x=2x = 2: y=233(22)+3=812+3=1y = 2^3 - 3(2^2) + 3 = 8 - 12 + 3 = -1. This is a local minimum.
  2. Inflection Points: To find points of inflection, calculate the second derivative: y=d2ydx2=6x6.y'' = \frac{d^2y}{dx^2} = 6x - 6.

    Solve y=0y'' = 0: 6x6=0x=1.6x - 6 = 0 \quad \Rightarrow \quad x = 1.

    At x=1x = 1, test the concavity:

    • For x<1x < 1 (e.g., x=0x = 0): y=6(0)6=6<0y'' = 6(0) - 6 = -6 < 0 → Concave down.
    • For x>1x > 1 (e.g., x=2x = 2): y=6(2)6=6>0y'' = 6(2) - 6 = 6 > 0 → Concave up.

    Therefore, x=1x = 1 is an inflection point. Substituting x=1x = 1 into y(x)y(x): y=133(12)+3=13+3=1.y = 1^3 - 3(1^2) + 3 = 1 - 3 + 3 = 1. Inflection point: (1,1)(1, 1).


c) Sketch the graph

The graph should include:

  • Critical points: (0,3)(0, 3) (max), (2,1)(2, -1) (min).
  • Inflection point: (1,1)(1, 1).
  • Intervals of increase/decrease and concavity.

I'll plot this if you'd like.


Would you like me to plot the graph or explain further?

Related Questions:

  1. How do you analyze concavity using the second derivative?
  2. Why do critical points help in identifying maxima and minima?
  3. What is the significance of inflection points in curve behavior?
  4. Can you find the x-intercepts of y=x33x2+3y = x^3 - 3x^2 + 3?
  5. How does increasing/decreasing behavior relate to derivatives?

Tip:

Always double-check the sign changes of the derivative to avoid mistakes when identifying increasing or decreasing intervals.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Critical Points
Intervals of Increase and Decrease
Inflection Points

Formulas

First derivative: y' = 3x^2 - 6x
Second derivative: y'' = 6x - 6

Theorems

Derivative Test for Increasing/Decreasing
Second Derivative Test for Concavity and Inflection Points

Suitable Grade Level

Grades 11-12