The answer is A... I'm stuck, should i foil 4x + 2 and x - 1????

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- Jul 17th 2013, 10:53 PMasilvester635Help on a question
The answer is A... I'm stuck, should i foil 4x + 2 and x - 1????

- Jul 17th 2013, 11:27 PMchiroRe: Help on a question
Hey asilvester635.

Hint: f(g(x)) = 2 * ([x-1]/2) + 1 = 2u + 1 where u = [x-1]/2. - Jul 18th 2013, 12:38 AMasilvester635Re: Help on a question
Thank you!!!

- Jul 18th 2013, 02:19 AMProve ItRe: Help on a question
It appears that you have thought $\displaystyle \displaystyle \begin{align*} f\left( g(x) \right) \end{align*}$ means $\displaystyle \displaystyle \begin{align*} f \cdot g(x) \end{align*}$, and it's not surprising you might think that, as we often use letters or numbers outside brackets to denote multiplication. However, in function notation, we have another meaning.

Notice that $\displaystyle \displaystyle \begin{align*} f(x) \end{align*}$ means "f as a function of x". A function can be thought of like a computer program, where you would put a number in and get a single number out, depending on what process your function does to the numbers. So we sometimes like to write $\displaystyle \displaystyle \begin{align*} x \to f \end{align*}$, which means we put in a number "x" and when the function does its job, we get out a number "f". So $\displaystyle \displaystyle \begin{align*} f(x) \end{align*}$ is a notation we use to show what is going in (the x) and what is coming out (the f).

When we have $\displaystyle \displaystyle \begin{align*} f \left( g(x) \right) \end{align*}$, this is a**composition of functions**, sometimes written as $\displaystyle \displaystyle \begin{align*} f \circ g(x) \end{align*}$. What's really happening is you have an input x going through a function and getting an output g, and then this output is going into another function to get the output f. So it can be thought of like they would say in Inception - a function in a function.

So when we have $\displaystyle \displaystyle \begin{align*} g(x) = \frac{x - 1}{2} \end{align*}$, that means when x goes through the function, it gets subtracted by 1 and then halved to get g. Then THIS will go through the function f. Since $\displaystyle \displaystyle \begin{align*} f(x) = 2x + 1 \end{align*}$, whatever gets input is doubled then has 1 added to it.

Therefore $\displaystyle \displaystyle \begin{align*} f\left( g(x) \right) \end{align*}$ means "use the output of g as the input of f and see what comes out". So if we put in $\displaystyle \displaystyle \begin{align*} \frac{x - 1}{2} \end{align*}$ and do what the function f tells it to do (double and add 1), then we get

$\displaystyle \displaystyle \begin{align*} f \left( g(x) \right) = 2 \left( \frac{x - 1}{2} \right) + 1 \end{align*}$.

Hope that helped.