If \displaystyle f(x)=\sqrt{x^2-2x+1}, what is \displaystyle f(9)?

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TheoryPractice · Accepted Answer

8. Function questions tend to derive most of their difficulty from the abstract function notation itself. So being comfortable with approaching function notation is most of the battle. When you see function notation such as \displaystyle f(x), keep in mind that \displaystyle x is the "input" (whatever they tell you \displaystyle x is, put that into the equation), and that \displaystyle f(x) is the "output" (once you've put your input through the equation, the result is the value of \displaystyle f(x)).

So when you're given \displaystyle f(x)=\sqrt{x^2-2x+1}, what the problem is really saying is that "whatever we put in the parentheses of \displaystyle f(), plug that value in wherever you see an \displaystyle x in \displaystyle f(x)=\sqrt{x^2-2x+1}.

Which means you'll take the input value, square it, subtract the product of the input value and two, add one to that, and then take the square root of the whole thing. With 9, that looks like:

\displaystyle f(9)=\sqrt{9^2-2(9)+1}

You can then simplify the math underneath the radical to get:

\displaystyle \sqrt{81-18+1}=\sqrt{64}

And since you know that \displaystyle \sqrt{64}=8 you have your answer.

If <tex>\displaystyle f(x)=\sqrt{x^2-2x+1}</tex>, what is <tex>\displaystyle f(9)</tex>?

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