Question 178998
"Given f(x)=1/[x], find f(2)+f(-2)"


Is the function {{{f(x)=1/abs(x)}}}? (ie is the right side 1 over the absolute value of "x")???



If so, then...



Let's find f(2)


{{{f(x)=1/abs(x)}}} Start with the given function



{{{f(2)=1/abs(2)}}} Plug in {{{x=2}}}



{{{f(2)=1/2}}} Take the absolute value of 2 to get 2



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Let's find f(-2)


{{{f(x)=1/abs(x)}}} Start with the given function



{{{f(-2)=1/abs(-2)}}} Plug in {{{x=-2}}}



{{{f(-2)=1/2}}} Take the absolute value of -2 to get 2



Now let's add f(2) to f(-2):



{{{f(2)+f(-2)=1/2+1/2=(1+1)/2=2/2=1}}}



So {{{f(2)+f(-2)=1}}}



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"Given f(x)=1/[x], find f(c^2+4)"



{{{f(x)=1/abs(x)}}} Start with the given function



{{{f(c^2+4)=1/abs(c^2+4)}}} Plug in {{{x=c^2+4}}}



From here, there's not much you can do to simplify.