Question 216335
{{{f(x)=2x^2+3x-5}}}
{{{g(x)=4}}}<br>
To solve for fog (pronounced "f of g") and gof ("pronounced g of f") it helps to understand function notation well. When you are given a formula (or rule) for a function (like {{{f(x)=2x^2+3x-5}}}) it is important to understand the role of "x". It is simply a place-holder. It represents whatever input you provide to the function. And the rule illustrates what the particular function will do with its input in determining the output for that input.<br>
So when you see {{{f(x)=2x^2+3x-5}}} it shows you what that function f will do with its input, "x". It will square it, multiply that by 2, add 3 times the input and then subtract 5. The function f will do this to <b>any</b> input you give it.
{{{f(x)=2x^2+3x-5}}}
{{{f(q)=2q^2+3q-5}}}
{{{f(z)=2z^2+3z-5}}}
{{{f(9)=2(9)^2+3(9)-5}}}
{{{f(4x+3)=2(4x+3)^2+3(4x+3)-5}}}
For {{{g(x) = 4}}}, since there is no "x" in the rule, the function g ignores the input! It simply returns a 4 no matter what input is given to the function!
{{{g(y) = 4}}}
{{{g(q) = 4}}}
{{{g(x^2-3x+99) = 4}}}
etc.
<br>
So now we are in position to figure out your problem. fog ("f of g") is another way of saying f(g(x)). Since g(x) = 4, f(g(x)) = f(4). And we saw above what f does to its input. It squares it, multiplies that by 2, etc. This is what it will do to 4:
{{{f(4)=2(4)^2+3(4)-5}}}
Now all we have to do is go through the Order of Operations (aka PEMDAS) to simplify it:
{{{f(4)=2(16)+3(4)-5}}}
{{{f(4)=32+12-5}}}
{{{f(4)=39}}}
So fog = f(g(x)) = f(4) = 39<br>
gof ("g of f") is another way of saying g(f(x)). Now, as we saw above, the g function totally ignores its input and simply returns 4 <i>all the time</i>. So gof = g(f(x)) = 4!