Solve y = -x² + 1: Finding Values Where Function is Positive

Question

Look at the following function:

y=x2+1 y=-x^2+1

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

f\left(x\right) > 0

Step-by-Step Solution

To solve the problem of finding the values of x x for which f(x)=x2+1>0 f(x) = -x^2 + 1 > 0 , we'll follow these steps:

  • Step 1: Begin with the inequality x2+1>0 -x^2 + 1 > 0 .
  • Step 2: Rearrange it to 1>x2 1 > x^2 .
  • Step 3: Recognize this can also be expressed as x2<1 x^2 < 1 .
  • Step 4: Since x2<1 x^2 < 1 , the solutions for x x are those values between 1-1 and 11: 1<x<1 -1 < x < 1 .

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

Answer

-1 < x < 1