Math Problem Statement

f(x) = (x^2 - 2x - 8)(x - 1). If 0 ≤ x ≤ 3/2, then find the range of values of x for which f(x) ≥ 0.

Solution

The function given is f(x)=(x22x8)(x1)f(x) = (x^2 - 2x - 8)(x - 1), and the goal is to find the range of values of xx for which f(x)0f(x) \geq 0 given that 0x320 \leq x \leq \frac{3}{2}.

Step 1: Factorize the quadratic expression

First, we factor the quadratic expression x22x8x^2 - 2x - 8.

We can factor it as: x22x8=(x4)(x+2)x^2 - 2x - 8 = (x - 4)(x + 2)

So, the function becomes: f(x)=(x4)(x+2)(x1)f(x) = (x - 4)(x + 2)(x - 1)

Step 2: Find the roots of the function

The function will be zero when any of the factors are zero. The roots are:

  • x4=0x=4x - 4 = 0 \Rightarrow x = 4
  • x+2=0x=2x + 2 = 0 \Rightarrow x = -2
  • x1=0x=1x - 1 = 0 \Rightarrow x = 1

These roots divide the real line into different intervals. We are interested in the behavior of the function within the interval [0,32][0, \frac{3}{2}].

Step 3: Analyze the sign of the function in [0,32][0, \frac{3}{2}]

We now check the sign of f(x)f(x) within the given interval:

  • At x=0x = 0, f(0)=(04)(0+2)(01)=(4)(2)(1)=8f(0) = (0 - 4)(0 + 2)(0 - 1) = (-4)(2)(-1) = 8 (positive).
  • At x=1x = 1, f(1)=(14)(1+2)(11)=(3)(3)(0)=0f(1) = (1 - 4)(1 + 2)(1 - 1) = (-3)(3)(0) = 0.
  • At x=32x = \frac{3}{2}, f(32)=(324)(32+2)(321)f\left(\frac{3}{2}\right) = \left(\frac{3}{2} - 4\right)\left(\frac{3}{2} + 2\right)\left(\frac{3}{2} - 1\right).

Let's calculate f(32)f\left(\frac{3}{2}\right):

f(32)=(52)(72)(12)=(358)f\left(\frac{3}{2}\right) = \left(-\frac{5}{2}\right)\left(\frac{7}{2}\right)\left(\frac{1}{2}\right) = \left(-\frac{35}{8}\right) which is negative.

Step 4: Conclusion

The function f(x)0f(x) \geq 0 between 00 and 11. For values of xx in (1,32]\left( 1, \frac{3}{2} \right], the function becomes negative.

Thus, the range of values of xx for which f(x)0f(x) \geq 0 in the interval 0x320 \leq x \leq \frac{3}{2} is: x[0,1]x \in [0, 1]

Would you like further details on any step? Here are five related questions to expand on this:

  1. How can you verify the behavior of f(x)f(x) for other intervals outside [0,32][0, \frac{3}{2}]?
  2. What role do critical points play in finding the sign of a function?
  3. How do you determine the concavity of a function like f(x)f(x)?
  4. Can you extend this problem to find where f(x)>0f(x) > 0?
  5. How would the analysis change if the interval for xx were different?

Tip: When solving inequalities involving factored polynomials, it is useful to evaluate the sign of each factor across the intervals formed by the roots.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Inequalities

Formulas

Quadratic Factorization
Polynomial Sign Analysis

Theorems

Zero Product Property
Sign Analysis for Polynomials

Suitable Grade Level

Grades 9-11