What is the domain of the function y = √(2x² - 5x - 12)?

Steps to Solve

1. Set the expression under the square root greater than or equal to zero

To find the domain of the function ( y = \sqrt{2x^2 - 5x - 12} ), we need to solve the inequality:

$$ 2x^2 - 5x - 12 \geq 0 $$

2. Factor the quadratic expression

Next, we will factor the quadratic expression. We need to find two numbers that multiply to ( 2 \times -12 = -24 ) and add to ( -5 ). The factors are ( -8 ) and ( 3 ). We can rewrite the expression as:

$$ 2x^2 - 8x + 3x - 12 = 0 $$

Now, we can factor by grouping:

$$ 2x(x - 4) + 3(x - 4) = 0 $$

Thus, we have:

$$ (2x + 3)(x - 4) = 0 $$

3. Find the critical points

Setting each factor to zero gives us the critical points:

$$ 2x + 3 = 0 \implies x = -\frac{3}{2} $$

$$ x - 4 = 0 \implies x = 4 $$

4. Sign analysis

Now, we will test intervals defined by these critical points ( (-\infty, -\frac{3}{2}) ), ( (-\frac{3}{2}, 4) ), and ( (4, \infty) ) to determine where the expression is non-negative.

• For ( x < -\frac{3}{2} ), choose ( x = -2 ):

$$ 2(-2)^2 - 5(-2) - 12 = 8 + 10 - 12 = 6 ; (> 0) $$

• For ( -\frac{3}{2} < x < 4 ), choose ( x = 0 ):

$$ 2(0)^2 - 5(0) - 12 = -12 ; (< 0) $$

• For ( x > 4 ), choose ( x = 5 ):

$$ 2(5)^2 - 5(5) - 12 = 50 - 25 - 12 = 13 ; (> 0) $$

5. Write the solution in interval notation

The quadratic expression ( 2x^2 - 5x - 12 ) is non-negative in the intervals ( (-\infty, -\frac{3}{2}] ) and ( [4, \infty) ). Therefore, the domain of the function is:

$$ (-\infty, -\frac{3}{2}] \cup [4, \infty) $$