solve the following inequality.
(x-3)^2/(x^2-4)is greater than or equal to 0.
sketch y = (x-3)^2 /((x-2)(x+2) )
your solution will be all the x's when the graph is above the x-axis
I can see it there is a zero at x = 3
and there are 2 vertical asymptotes at
x = 2 , and x = -2
How about thinking about this intuitively
the numerator is always ≥ 0, since we squared something
Also we can see that x^2 - 4 is negative for values of x between -2 and 2, and x cannot be +2 or -2
so it looks like the expression is above the x-axis for x > 2 OR x < -2
how do you write that in interval notation?
To solve the inequality `(x-3)^2 / (x^2-4) ≥ 0`, we need to find the values of x that make the expression greater than or equal to zero.
Step 1: Find the critical points
The critical points are the values of x that make the numerator or denominator equal to zero.
Numerator: (x-3)^2 = 0
This occurs only when x = 3
Denominator: x^2 - 4 = 0
This equation can be factored as (x-2)(x+2) = 0
Hence, x = 2 or x = -2
Step 2: Create a number line
We divide the number line into intervals using these critical points: -∞, -2, 2, 3, +∞
Step 3: Choose test points in each interval
Select a test point from each interval created in Step 2 and evaluate the inequality `(x-3)^2 / (x^2-4) ≥ 0`.
Interval 1: x < -2
Choose x = -3: (x-3)^2 / (x^2-4) = (-3-3)^2 / ((-3)^2-4) = 36 / 5 > 0
The inequality is true in this interval.
Interval 2: -2 < x < 2
Choose x = 0: (x-3)^2 / (x^2-4) = (0-3)^2 / (0^2-4) = 9 / -4 < 0
The inequality is false in this interval.
Interval 3: 2 < x < 3
Choose x = 2.5: (x-3)^2 / (x^2-4) = (2.5-3)^2 / (2.5^2-4) = 0
The inequality is true on this interval.
Interval 4: x > 3
Choose x = 4: (x-3)^2 / (x^2-4) = (4-3)^2 / (4^2-4) = 1 / 12 > 0
The inequality is true in this interval.
Step 4: Analyze the intervals
From our test point evaluations, we can see that the inequality `(x-3)^2 / (x^2-4) ≥ 0` is true in the intervals x < -2, 2 < x < 3, and x > 3.
Step 5: Write the solution
The solution to the inequality `(x-3)^2 / (x^2-4) ≥ 0` is:
x < -2, 2 < x < 3, x > 3