Solving x²-1/4: Finding Negative Function Values and Domains

Q: Why do I need to find where the function equals zero first?

The zeros are boundary points where the function changes from positive to negative (or vice versa). For $ y = x^2 - \frac{1}{4} $, these occur at $ x = \pm\frac{1}{2} $ and divide the number line into intervals.

Q: How do I know the function is negative between the roots?

Since this is a parabola opening upward (positive $ x^2 $ coefficient), it dips below the x-axis between its roots. Test any value like $ x = 0 $: \( f(0) = -\frac{1}{4} ✓

Q: What if I can't solve x² = 1/4 easily?

Remember that $ x^2 = \frac{1}{4} $ means $ x = \pm\sqrt{\frac{1}{4}} = \pm\frac{1}{2} $. The square root of a fraction is the square root of numerator over square root of denominator!

Q: Why is the answer an open interval with < symbols?

We want $ f(x) (strictly less than), not \( f(x) ≤ 0 $. At $ x = \pm\frac{1}{2} $, the function equals zero, so we use open interval notation $ (-\frac{1}{2}, \frac{1}{2}) $.

Q: How do I remember which interval is negative?

Think of the parabola shape! Since it opens upward and crosses the x-axis at $ x = -\frac{1}{2} $ and $ x = \frac{1}{2} $, the function is negative in the valley between these points.

Quadratic Functions with Sign Analysis

Find the positive and negative domains of the function below:

$y=x^2-\frac{1}{4}$

Then determine for which values of $x$ the following is true:

$f(x) < 0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the positive and negative domains of the function below:

$y=x^2-\frac{1}{4}$

Then determine for which values of $x$ the following is true:

$f(x) < 0$

Step-by-step solution

To solve this problem, we will follow these steps:

Step 1: Solve the equation $x^2 = \frac{1}{4}$ to find critical points.
Step 2: Determine the sign of $f(x) = x^2 - \frac{1}{4}$ in each interval between and beyond these points.
Step 3: Identify the interval where $f(x) < 0$ .

Now, let's work through each step:
Step 1: Solve the equation $x^2 = \frac{1}{4}$ . The solutions are $x = \frac{1}{2}$ and $x = -\frac{1}{2}$ because solving this gives $x = \pm \sqrt{\frac{1}{4}}$ .
Step 2: To find where $f(x) < 0$ , we need $x^2 < \frac{1}{4}$ . This is equivalent to finding where $-\frac{1}{2} < x < \frac{1}{2}$ . In this interval, $f(x)$ is negative.
Step 3: The function $y = x^2 - \frac{1}{4}$ is negative for values of $x$ between these roots, i.e., for $-\frac{1}{2} < x < \frac{1}{2}$ .

Therefore, the solution is that the function is negative for $-\frac{1}{2} < x < \frac{1}{2}$ .

Final Answer

$-\frac{1}{2} < x < \frac{1}{2}$

Key Points to Remember

Essential concepts to master this topic

Critical Points: Set function equal to zero to find boundary values
Technique: Test values in intervals: $f(0) = 0^2 - \frac{1}{4} = -\frac{1}{4} < 0$
Check: Verify endpoints: at $x = \pm\frac{1}{2}$ , function equals zero ✓

Common Mistakes

Avoid these frequent errors

Confusing where function is positive vs negative
Don't assume the function is negative everywhere just because it has a negative constant = wrong intervals! The parabola opens upward, so it's only negative between the roots. Always find the zeros first, then test values in each interval to determine signs.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below does not intersect the $$ x $$ -axis.

The parabola's vertex is marked A.

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

FAQ

Everything you need to know about this question

Why do I need to find where the function equals zero first?

The zeros are boundary points where the function changes from positive to negative (or vice versa). For $y = x^2 - \frac{1}{4}$ , these occur at $x = \pm\frac{1}{2}$ and divide the number line into intervals.

How do I know the function is negative between the roots?

Since this is a parabola opening upward (positive $x^2$ coefficient), it dips below the x-axis between its roots. Test any value like $x = 0$ : $f(0) = -\frac{1}{4} < 0$ ✓

What if I can't solve x² = 1/4 easily?

Remember that $x^2 = \frac{1}{4}$ means $x = \pm\sqrt{\frac{1}{4}} = \pm\frac{1}{2}$ . The square root of a fraction is the square root of numerator over square root of denominator!

Why is the answer an open interval with < symbols?

We want $f(x) < 0$ (strictly less than), not $f(x) ≤ 0$ . At $x = \pm\frac{1}{2}$ , the function equals zero, so we use open interval notation $(-\frac{1}{2}, \frac{1}{2})$ .

How do I remember which interval is negative?

Think of the parabola shape! Since it opens upward and crosses the x-axis at $x = -\frac{1}{2}$ and $x = \frac{1}{2}$ , the function is negative in the valley between these points.

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