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by **jdom543** on Sat Sep 04, 2010 3:01 pm

$f(x)=\frac{1-cos(x)}{x}$ and $g(x)=\frac{2}{x}$ Interval: $(0,\infty )$

Show that $g(x)\ge f(x)\ge 0$

How do I do that?

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by **nona.m.nona** on Sat Sep 04, 2010 9:47 pm

jdom543 wrote: $f(x)=\frac{1-cos(x)}{x}$ and $g(x)=\frac{2}{x}$ Interval: $(0,\infty )$

Show that $g(x)\ge f(x)\ge 0$

For f(x): What do you know about the value of the cosine? Then what does this tell you about the value (and sign) of 1 - cos(x)? Given that x is always positive, what then do you know about the value of f(x)?

For g(x): The denominators are always the same, so you need only compare the numerators. What is the maximum value of the numerator of f(x)? How does this compare with the constant value of the numerator of g(x)?

The Purplemath Forums

Show g(x)>=f(x)>=0 for f(x)=(1-cosx)/x, g(x)=2/x

Show g(x)>=f(x)>=0 for f(x)=(1-cosx)/x, g(x)=2/x