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state the domain (rounded accurate to four decimal places) and the range 9exact values) of the function f(x)= (x^3-5x+2)^1/2.

It is (x^3-5x+2) under square root.
Thanks

2007-09-22 07:27:50 · 2 answers · asked by Kakak 1 in Science & Mathematics Mathematics

and the range (exact values) ignore the 9

2007-09-22 07:31:33 · update #1

2 answers

f(x)= (x^3-5x+2)^1/2.
Domain= [-2.4142, .4142] U [2, =infinity)
Range is y >= 0

x^3-5x+2 = (x-2)(x^2+ 2x- 1)
(x^2+2x-1) has roots = -1 +/- sqrt(2)

2007-09-22 08:00:27 · answer #1 · answered by ironduke8159 7 · 0 0

Well you know that you can't get the square root of a negative number. Further, if x < 0 then x^3 < 0. All you need to figure out is when is x^3 - 5x + 2 > 0 to get the domain.

As for the range, the square root of any number has to be positive (or 0). Also, as x approaches infinity, f(x) will approach infinity. The range, then, becomes easy. y | y >= 0

2007-09-22 07:35:42 · answer #2 · answered by gitter1226 5 · 0 0

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