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December 17, 2014

December 17, 2014

Posted by **Jordan** on Monday, October 7, 2013 at 8:29pm.

a)f(x)=(x^2+3)^2,XER

Answer is f(x)ER, f(x)>9

b)f(x)=1/3+x^4,XER

Answer is f(x)ER, 0<f(x)<1/3

Please explain how they go those two answers

- Math -
**Steve**, Tuesday, October 8, 2013 at 12:32amx^2+3 >= 3 for all x

so, f(x) >= 9

3+x^4 >= 3 for all x

1/(3+x^4) <= 1/3 for all x

as x gets huge, 1/x^4 -> 0 so,

0 < f(x) <= 1/3

wolframalpha.com will graph them for you to see

- Math -
**Jai**, Tuesday, October 8, 2013 at 12:37amI suppose that "XER" means the domain (all possible values of x) is all real numbers. And the answers there are both...wrong.

Anyway, range is the set of all possible values of y (or f(x)). Let's analyze each problem. :)

a) f(x) = (x^2+3)^2

Note that x^2 can never be negative, because any real number squared is always greater than or equal to zero. Thus the smallest possible value of x^2 is zero, and if we substitute it in f(x),

f(x) = (0^2 + 3)^2

f(x) = 3^2

f(x) = 9, which is the smallest possible value of f(x)

Thus the range is all real numbers greater than or equal to 9, or f(x) >= 9

b) f(x) = 1/3+x^4

x^4 is also (x^2)^2. Then again, x^4 can never be negative. So the smallest possible value of x^4 is zero, and substituting,

f(x) = (1/3) + 0^4

f(x) = 1/3

Thus the range is all real numbers greater than or equal to 1/3, or f(x) >= 1/3

Hope this helps~ :3

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