Domain Analysis: Finding Where (x-1/3)(-x-2¼) > 0

Quadratic Inequalities with Mixed Number Zeros

Find the positive and negative domains of the function:

y=(x13)(x214) y=\left(x-\frac{1}{3}\right)\left(-x-2\frac{1}{4}\right)

Determine for which values of x x the following is true:

f(x)>0 f(x) > 0

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Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find the positive and negative domains of the function:

y=(x13)(x214) y=\left(x-\frac{1}{3}\right)\left(-x-2\frac{1}{4}\right)

Determine for which values of x x the following is true:

f(x)>0 f(x) > 0

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Identify the roots of the quadratic expression.
  • Step 2: Determine the sign of each factor in intervals defined by these roots.
  • Step 3: Identify where the product of these factors is positive.

Now, let's work through each step:

Step 1: Identify the roots.
The given function is y=(x13)(x214) y = \left(x - \frac{1}{3}\right)\left(-x - 2\frac{1}{4}\right) . To find the roots, solve each factor for zero:

  • x13=0 x - \frac{1}{3} = 0 gives x=13 x = \frac{1}{3} .
  • x214=0-x - 2\frac{1}{4} = 0 gives x=214 x = -2\frac{1}{4} .

Step 2: Determine the sign of each factor in the intervals defined by these roots.
The zeros divide the x-axis into three intervals: (,214)(-∞, -2\frac{1}{4}), (214,13)(-2\frac{1}{4}, \frac{1}{3}), and (13,)(\frac{1}{3}, ∞).

Step 3: Test the signs and find where the product is positive.

  • For x<214 x < -2\frac{1}{4} , test with x=3 x = -3 : - x13=313<0 x - \frac{1}{3} = -3 - \frac{1}{3} < 0 - x214=3214>0-x - 2\frac{1}{4} = 3 - 2\frac{1}{4} > 0 - Product: Negative
  • For 214<x<13 -2\frac{1}{4} < x < \frac{1}{3} , test with x=0 x = 0 : - x13=013<0 x - \frac{1}{3} = 0 - \frac{1}{3} < 0 - x214=0214>0-x - 2\frac{1}{4} = 0 - 2\frac{1}{4} > 0 - Product: Positive
  • For x>13 x > \frac{1}{3} , test with x=1 x = 1 : - x13=113>0 x - \frac{1}{3} = 1 - \frac{1}{3} > 0 - x214=1214<0-x - 2\frac{1}{4} = -1 - 2\frac{1}{4} < 0 - Product: Negative

Therefore, the solution to f(x)>0 f(x) > 0 is in the interval:

214<x<13 -2\frac{1}{4} < x < \frac{1}{3} .

3

Final Answer

214<x<13 -2\frac{1}{4} < x < \frac{1}{3}

Key Points to Remember

Essential concepts to master this topic
  • Find Zeros: Set each factor equal to zero to find critical points
  • Test Intervals: Use x=0 x = 0 in middle interval: negative × negative = positive
  • Sign Analysis: Check product signs in all three intervals around zeros ✓

Common Mistakes

Avoid these frequent errors
  • Testing only one interval instead of all three
    Don't just find the zeros and guess the answer = wrong solution! The function changes sign at each zero, so you must test every interval. Always check the sign of the product in all intervals between and outside the zeros.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of \( x \) where \( f\left(x\right) > 0 \).

AAABBBCCCX

FAQ

Everything you need to know about this question

Why do I need to test points in each interval?

+

The function changes sign at each zero! Testing points tells you whether the product is positive or negative in each interval between the zeros.

How do I handle the mixed number in the second factor?

+

Convert 214 2\frac{1}{4} to an improper fraction: 94 \frac{9}{4} . So x214=0 -x - 2\frac{1}{4} = 0 becomes x=94 x = -\frac{9}{4} .

What if I get confused about which intervals are positive?

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Make a sign chart! Draw a number line with your zeros, then test one point in each interval. Mark each interval as + or - based on your test results.

Why is the answer between the two zeros?

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When you have two factors with opposite signs in the middle interval, their product becomes positive! Outside this interval, both factors have the same sign, making the product negative.

Do I include the zeros in my final answer?

+

No! The inequality is f(x)>0 f(x) > 0 (strictly greater than), so the zeros where f(x)=0 f(x) = 0 are not included in the solution.

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