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http://assets.openstudy.com/updates/attachments/53a5a672e4b0a819ab10981e-jenniferjuice-1403365232824-....png Given that f(x) = x2 − 3x + 3 and g(x) = the quantity of x minus one, over four, solve for f(g(x)) when x = 5. −3 1 7 13
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Plug in g(x) into f( ) \(f(\color{red}{x})= \color{red}{x}^2-3\color{red}{x}+3\), So the idea will be to substitute g(x) wherever you see "x" in your function: \(f(\color{red}{g(x)}) =(\color{red}{g(x)})^2-3\color{red}{g(x)}+3 \) \[ f\left(\color{red}{\frac{x-1}{4} }\right)=\left(\color{red}{\frac{x-1}{4} }\right)^2-3\left(\color{red}{\frac{x-1}{4} }\right)+3\]
Then all you need to do is substitute the value of x=5 in the last expression above^
is it 1 ?
yes indeed
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