Functions - Set of Values

Jul 2010
7
0
1. Let \(\displaystyle f(x)=$\sqrt{(1/x^2)-2)}$\)

Find

a) The set of real values of x for which f is real and finite

b) The range of f

2.

Let \(\displaystyle f(x)=(x+4)/(x+1)\) , x ≠ -1 and \(\displaystyle g(x)=(x-2)/(x-4)\) x ≠ 4

Find the set of values of x such that f(x) ≤ g(x)

Please help. Thanks! :)
 
Nov 2008
1,458
646
France
Hi

\(\displaystyle f(x)=$\sqrt{(1/x^2)-2)}$\)

f is defined when \(\displaystyle (1/x^2)-2 \geq 0\)


Let \(\displaystyle f(x)=(x+4)/(x+1)\) , x ≠ -1 and \(\displaystyle g(x)=(x-2)/(x-4)\) x ≠ 4

\(\displaystyle g(x) \geq f(x)\)

\(\displaystyle \displaystyle \frac{x-2}{x-4} - \frac{x+4}{x+1} \geq 0\)

Use the same denominator to simplify this expression
 
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