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Find the domain and range of function?

0 votes
Function h(x) = sqrt2x^2+3x+1
asked Oct 21, 2014 in ALGEBRA 2 by anonymous

1 Answer

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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}.

answered Oct 21, 2014 by casacop Expert
edited Oct 21, 2014 by casacop

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