Calculate Positive Area: Finding Area of f(x)=2x²-50

Quadratic Inequalities with Sign Analysis

Find the positive area of the function

f(x)=2x250 f(x)=2x^2-50

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find the positive domain of the function
00:03 Looking at the coefficient of X squared, which is positive, therefore the function is smiling
00:11 The positive domain is actually above the X-axis
00:14 Therefore, we'll set Y=0 to find the intersection points with the X-axis
00:21 Isolate X
00:32 Extract the root
00:38 When extracting a root, there are 2 solutions (positive and negative)
00:43 These are the intersection points with the X-axis
00:56 Let's mark the intersection points with the X-axis
01:06 The positive domain is above the X-axis
01:10 The negative domain is below the X-axis
01:14 Let's find the domain where the function is positive
01:21 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find the positive area of the function

f(x)=2x250 f(x)=2x^2-50

2

Step-by-step solution

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

  • Step 1: Solve the equation 2x250=0 2x^2 - 50 = 0 to find the roots.
  • Step 2: Analyze intervals on the number line based on the roots to find where f(x)>0 f(x) > 0 .

Now, let's work through each step:
Step 1: Set the function to zero: 2x250=0 2x^2 - 50 = 0 . Solving gives:

2x2=50x2=25x=±5 2x^2 = 50 \\ x^2 = 25 \\ x = \pm 5

Step 2: The roots divide the number line into intervals: x<5 x < -5 , 5<x<5 -5 < x < 5 , and x>5 x > 5 . We test these intervals in f(x)=2x250 f(x) = 2x^2 - 50 .

  • For x<5 x < -5 , pick x=6 x = -6 : f(6)=2(6)250=7250=22>0 f(-6) = 2(-6)^2 - 50 = 72 - 50 = 22 > 0 This interval is positive.
  • For 5<x<5 -5 < x < 5 , pick x=0 x = 0 : f(0)=2(0)250=50<0 f(0) = 2(0)^2 - 50 = -50 < 0 This interval is negative.
  • For x>5 x > 5 , pick x=6 x = 6 : f(6)=2(6)250=7250=22>0 f(6) = 2(6)^2 - 50 = 72 - 50 = 22 > 0 This interval is positive.

Therefore, the function is positive for x<5 x < -5 and x>5 x > 5 .

Hence, the solution to the problem is x<5 x < -5 or 5<x 5 < x .

3

Final Answer

x<5 x < -5 o 5<x 5 < x

Key Points to Remember

Essential concepts to master this topic
  • Root Finding: Set quadratic equal to zero to find boundary points
  • Interval Testing: Pick test values like x = -6, 0, 6 in each interval
  • Verification: Check that f(-6) = 22 > 0 and f(6) = 22 > 0 ✓

Common Mistakes

Avoid these frequent errors
  • Testing only one interval or guessing the sign pattern
    Don't assume the parabola is positive everywhere outside the roots without testing = wrong intervals! The sign can change unpredictably. Always test a specific value in each interval created by the roots.

Practice Quiz

Test your knowledge with interactive questions

Find the ascending area of the function

\( f(x)=2x^2 \)

FAQ

Everything you need to know about this question

Why do I need to find the roots first if I'm looking for positive areas?

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The roots are boundary points where the function changes sign! For f(x)=2x250 f(x) = 2x^2 - 50 , the roots x = ±5 divide the number line into intervals where the function is consistently positive or negative.

How do I know which test values to pick in each interval?

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Pick any convenient number in each interval! For x<5 x < -5 , try x = -6. For 5<x<5 -5 < x < 5 , try x = 0. For x>5 x > 5 , try x = 6. The specific value doesn't matter as long as it's in the right interval.

What does 'positive area' actually mean?

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Positive area means where the function is above the x-axis (f(x) > 0). We're finding the x-values where the parabola is positive, not calculating an actual area measurement.

Why is the middle interval negative when it's between the roots?

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This parabola f(x)=2x250 f(x) = 2x^2 - 50 opens upward (positive coefficient of x²). Between the roots, it dips below the x-axis, making f(x) negative. Outside the roots, it rises above the x-axis, making f(x) positive.

Can I just look at the graph instead of doing algebra?

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Graphing helps visualize, but algebraic methods give exact answers! Testing intervals ensures you get the precise boundaries (x<5 x < -5 or x>5 x > 5 ) rather than approximate values from a sketch.

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