Solve (x+1)(6-x): Finding Values Where Function is Positive

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

The roots are where the function changes sign! At $ x = -1 $ and $ x = 6 $, the function equals zero, and these points divide the number line into regions where the function is either positive or negative.

Q: How do I know which test points to choose?

Pick any convenient number from each interval! For $ (-\infty, -1) $, try $ x = -2 $. For $ (-1, 6) $, try $ x = 0 $. For $ (6, \infty) $, try $ x = 7 $. Any point in the interval will give the same sign!

Q: What's the difference between > and ≥ in the final answer?

Since we want $ f(x) > 0 $ (strictly greater than), we don't include the points where $ f(x) = 0 $. So our answer is $ -1 , not \( -1 ≤ x ≤ 6 $.

Q: Can I expand the expression first instead?

You could expand to get $ y = -x^2 + 5x + 6 $, but it's much harder to solve $ -x^2 + 5x + 6 > 0 $! The factored form makes it easy to see the roots and test intervals.

Q: What if I get the wrong sign when testing?

Double-check your arithmetic! For $ x = 0 $: $ (0+1)(6-0) = (1)(6) = 6 $. Make sure you're substituting correctly and following the order of operations carefully.

Quadratic Inequalities with Sign Analysis

Look at the following function:

$y=\left(x+1\right)\left(6-x\right)$

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

$f(x) > 0$

❤️ Continue Your Math Journey!

We have hundreds of course questions with personalized recommendations + Account 100% premium

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the following function:

$y=\left(x+1\right)\left(6-x\right)$

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

$f(x) > 0$

Step-by-step solution

To solve this problem, we follow these steps:

Step 1: Find the Roots
Set $(x+1)(6-x) = 0$ . Solving these linear equations, we get the roots:
- $x+1 = 0$ gives $x = -1$ .
- $6-x = 0$ gives $x = 6$ .
Step 2: Determine the Intervals
Based on the roots, the real line is divided into intervals: $(-\infty, -1)$ , $(-1, 6)$ , and $(6, \infty)$ .
Step 3: Test Each Interval for Sign
- For $x \in (-\infty, -1)$ : Choose $x = -2$ . The expression $(x+1)(6-x) = (-2+1)(6+2) = (-1)(8) = -8 < 0$ .
- For $x \in (-1, 6)$ : Choose $x = 0$ . The expression $(x+1)(6-x) = (0+1)(6-0) = 1 \cdot 6 = 6 > 0$ .
- For $x \in (6, \infty)$ : Choose $x = 7$ . The expression $(x+1)(6-x) = (7+1)(6-7) = 8(-1) = -8 < 0$ .

The solution, based on the interval where the product is positive, is when $-1 < x < 6$ .

Therefore, the values of $x$ for which $f(x) > 0$ are $-1 < x < 6$ .

Final Answer

$-1 < x < 6$

Key Points to Remember

Essential concepts to master this topic

Roots: Set each factor equal to zero to find critical points
Technique: Test intervals using $x = 0$ : $(0+1)(6-0) = 6 > 0$
Check: Verify boundary behavior: function equals zero at $x = -1$ and $x = 6$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting to test all intervals between roots
Don't just find the roots $x = -1$ and $x = 6$ and guess the answer = wrong solution! The sign can change at each root, so you might pick the wrong intervals. Always test a point in each interval to determine where the function is actually positive.

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 roots are where the function changes sign! At $x = -1$ and $x = 6$ , the function equals zero, and these points divide the number line into regions where the function is either positive or negative.

How do I know which test points to choose?

Pick any convenient number from each interval! For $(-\infty, -1)$ , try $x = -2$ . For $(-1, 6)$ , try $x = 0$ . For $(6, \infty)$ , try $x = 7$ . Any point in the interval will give the same sign!

What's the difference between > and ≥ in the final answer?

Since we want $f(x) > 0$ (strictly greater than), we don't include the points where $f(x) = 0$ . So our answer is $-1 < x < 6$ , not $-1 ≤ x ≤ 6$ .

Can I expand the expression first instead?

You could expand to get $y = -x^2 + 5x + 6$ , but it's much harder to solve $-x^2 + 5x + 6 > 0$ ! The factored form makes it easy to see the roots and test intervals.

What if I get the wrong sign when testing?

Double-check your arithmetic! For $x = 0$ : $(0+1)(6-0) = (1)(6) = 6$ . Make sure you're substituting correctly and following the order of operations carefully.

🌟 Unlock Your Math Potential

Get unlimited access to all 18 The Quadratic Function questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime

Solve (x+1)(6-x): Finding Values Where Function is Positive

Quadratic Inequalities with Sign Analysis

❤️ Continue Your Math Journey!

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

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

How do I know which test points to choose?

What's the difference between > and ≥ in the final answer?

Can I expand the expression first instead?

What if I get the wrong sign when testing?

🌟 Unlock Your Math Potential

More Questions

Product Representation

Solve (x+1)(6-x) > 0: Finding Positive Function Values

Solve (x-4)(-x+6) > 0: Finding Positive Value Ranges

Solve (3x+3)(2-x) > 0: Finding Positive Values of x

Solve (3x+1)(1-3x): Finding Values Where Function is Negative

Positive and Negative Domains

Solve (x-4)(-x+6): Finding Values Where Function is Positive

Solve (x+1)(x+5) > 0: Finding Positive Values of a Quadratic Function

Solve (3x+3)(2-x) < 0: Finding Values Where Function is Negative

Solve y = x² + 10x + 16: Finding Values Where Function is Positive