Math Problem Statement

f(x)=(x3)(x29)f\left(x\right)=\left(x-3\right)\left(x^2-9\right)

Solution

Let's work through the problem step by step for the function f(x)=(x3)(x29)f(x) = (x - 3)(x^2 - 9).

1. Expanding the Function

First, expand the function f(x)f(x):

f(x)=(x3)(x29)f(x) = (x - 3)(x^2 - 9)

Notice that x29x^2 - 9 is a difference of squares, so:

f(x)=(x3)(x3)(x+3)f(x) = (x - 3)(x - 3)(x + 3)

f(x)=(x3)2(x+3)f(x) = (x - 3)^2(x + 3)

Or if expanded completely:

f(x)=(x26x+9)(x+3)f(x) = (x^2 - 6x + 9)(x + 3)

Expanding this:

f(x)=x3+3x26x218x+9x+27f(x) = x^3 + 3x^2 - 6x^2 - 18x + 9x + 27

f(x)=x33x29x+27f(x) = x^3 - 3x^2 - 9x + 27

2. First Derivative f(x)f'(x)

Differentiate f(x)=x33x29x+27f(x) = x^3 - 3x^2 - 9x + 27 with respect to xx:

f(x)=ddx(x33x29x+27)f'(x) = \frac{d}{dx}(x^3 - 3x^2 - 9x + 27)

The derivatives of each term are:

f(x)=3x26x9f'(x) = 3x^2 - 6x - 9

3. Second Derivative f(x)f''(x)

Differentiate f(x)f'(x) to find the second derivative:

f(x)=ddx(3x26x9)f''(x) = \frac{d}{dx}(3x^2 - 6x - 9)

f(x)=6x6f''(x) = 6x - 6

4. Stationary Points

Stationary points occur where f(x)=0f'(x) = 0. Set f(x)=0f'(x) = 0:

3x26x9=03x^2 - 6x - 9 = 0

Divide the entire equation by 3:

x22x3=0x^2 - 2x - 3 = 0

Factorize the quadratic equation:

(x3)(x+1)=0(x - 3)(x + 1) = 0

Thus, x=3x = 3 and x=1x = -1 are the stationary points.

5. Nature of Stationary Points

Evaluate the second derivative f(x)f''(x) at the stationary points to determine their nature:

  • For x=3x = 3:

f(3)=6(3)6=186=12f''(3) = 6(3) - 6 = 18 - 6 = 12

Since f(3)>0f''(3) > 0, x=3x = 3 is a point of minimum.

  • For x=1x = -1:

f(1)=6(1)6=66=12f''(-1) = 6(-1) - 6 = -6 - 6 = -12

Since f(1)<0f''(-1) < 0, x=1x = -1 is a point of maximum.

6. Sketching the Graph

The graph will have:

  • A minimum point at (3,f(3))(3, f(3))
  • A maximum point at (1,f(1))(-1, f(-1))

To find the corresponding yy-values, substitute these xx-values back into the original function f(x)f(x).

  • f(3)=(33)(99)=0×0=0f(3) = (3 - 3)(9 - 9) = 0 \times 0 = 0
  • f(1)=(13)((1)29)=4(19)=4(8)=32f(-1) = (-1 - 3)((-1)^2 - 9) = -4(1 - 9) = -4(-8) = 32

So:

  • Minimum at (3,0)(3, 0)
  • Maximum at (1,32)(-1, 32)

Would you like any further details or a deeper explanation on any part of this solution?

Here are 5 questions to further explore:

  1. How does factorizing the expression help in determining the roots and stationary points?
  2. Why is the second derivative useful in determining the nature of stationary points?
  3. What would be the effect on the graph if the constant term in the expanded form was different?
  4. How can we use the first derivative test instead of the second derivative test to determine the nature of stationary points?
  5. How would the graph behave at very large positive and negative values of xx?

Tip:

When expanding or factoring polynomials, always recheck each step to avoid small errors that can lead to incorrect stationary points or graph shapes.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Polynomials
Quadratic Equations
Derivatives
Stationary Points

Formulas

Difference of squares
Quadratic formula

Theorems

Second derivative test

Suitable Grade Level

Advanced High School