Find the domain and range of function?

Question

Function h(x) = sqrt2x^2+3x+1

Answer 1

The function is h(x) = √(2x²+3x+1)

y = √(2x²+3x+1)

The domain of a function is all values of x, those makes the function mathematically correct.

Since there shouldn't be any negative numbers in the square root.

√(2x²+3x+1) ≥ 0

2x²+3x+1 = 0

(2x+1)(x+1) = 0

x = -1/2, -1.

Therefore the domain of h(x) is {x Є R: x ≤ -1 or x ≥ -1/2}

Range set is the corresponding values of the function for different values of x.

Since there shouldn't be any negative numbers in the square root.

Therefore the range of the function is {y Є R: y ≥ 0}

Therefore the domain of h(x) is {x Є R: x ≤ -1 or x ≥ -1/2} and range is {y Є R: y ≥ 0}.