Given an equation for a function, e.g. \(y=x^2\) , or \(y=\dfrac{x}{x+1}\) we can let \(f(x)\) replace the expression in \(x\) appearing to the right of the equal sign. In this way, we can speak about functions in general as being of the form \(y=f(x)\) , without having to specify the particular form of the expression in \(x\). Furthermore, this allows us to name a function using a letter. Thus, if we speak of the function \(g\), we mean that there is some mathematical expression \(g(x)\) and an equation \(y=g(x)\) where \(g(x)\) gives the instructions on how to compute the output number \(y\) from the input number \(x\).

For example, if we speak of the function \(h(x)=\dfrac{x^2+1}{2}\), then we know that the equation of the function is \(y=\dfrac{x^2+1}{2}\), and we know that we compute the output number \(y\) corresponding to a given input number \(x\) by adding one to the square of the input number \(x\), then dividing by two.

Now, one might ask, if we already have \(h(x)=\dfrac{x^2+1}{2}\), why do we need to write \(y=\dfrac{x^2+1}{2}\)? The answer is: **we dont have to**. We can think
of \(h(x)\) itself as representing the output number corresponding to the input number when \(x\) is put into the function \(h\). Now keep in mind that we are now thinking
of \(h(x)\) in two different ways. On the one hand we are thinking of \(h(x)\) as representing the **mathematical expression** \(\dfrac{x^2+1}{2}\), and on the other hand we
are regarding \(h(x)\) as a symbol for the **numerical output** of the function \(h\) when \(x\) is the input. So, for example, \(h(3)\) stands for the expression
\(\dfrac{3^2+1}{2}\) but \(h(3)\) also stands for the number \(5\), which is the **value** of the expression \(\dfrac{3^2+1}{2}\).

Make a table containing two columns. Label the first column \(x\) and the second column \(h(x)\). Pick five values for \(x\) and place them in the first column. Pick some decimal values as well as whole number values, and negative as well as positive values. Pick all input values of \(x\) such that their absolute values are not larger than 2. Now, compute the corresponding output values of \(h(x)\) using \(h(x)=\dfrac{x^2+1}{2}\) and place the values in column 2. Plot each resulting pair of numbers \((x,y)\) in the Cartesian plane, where \(y=h(x)\). Graph Paper

Notice that \(h(2)=\dfrac{2^2+1}{2}\) and \(h(3)=\dfrac{3^2+1}{2}\), etc. So whatever number replaces \(x\) in the expression \(h(x)\) must replace each occurrence of \(x\) in the equation \(h(x)=\dfrac{x^2+1}{2}\). To make this more explicit, we could write the function \(h\) in blank parenthesis form as follows \(h(~~)=\dfrac{(~~)^2+1}{2}\). Then any number placed into the blank parenthesis following the \(h\) must also be placed into the blank parenthesis on the right side of the equation. The next example will show why this is an important idea.

Let us suppose that the input number \(x\) is itself computed from a number \(t\) according to the formula \(x=2t-1\). Then, since \(h(x)=\dfrac{x^2+1}{2}\),
it follows that \(h(2t-1)=\dfrac{(2t-1)^2+1}{2}=2t^2-2t+1\). Thus, not only can we place **numbers** into the blank parentheses in \(h(~~)=\dfrac{(~~)^2+1}{2}\), but we can
place **algebraic expressions** as well.

Find \(h(w+2)\), \(h(\sqrt{2x-1})\), and \(h(x+a)\). Simplify each expression.

See Solution

Find and simplify the value of \(f(x)=\dfrac{f(x+a)-f(x)}{a}\), given that \(f(x)=x^2\).

See Solution