College Algebra

Suppose $f(x)=2x+1$ and $g(x)=x^2$ and we want to find $f(g(-2))\text{.}$ Since we work from the inside out, that means we start by finding $g(-2)\text{.}$ We use the formula for $g(x)$ to get $g(-2)=(-2{)}^2=4\text{.}$ Now, we can plug that into $f(x)\text{:}$ $f(g(-2))=f(4)=2(4)+1=9\text{.}$ Therefore, our answer is $f(g(-2))=4\text{.}$

Now, let’s find $g(f(-2))$ instead. This time, we start with $f(x)\text{,}$ since it’s on the inside: $f(-2)=2(-2)+1=-3\text{.}$ Now, we can plug that into $g(x)\text{:}$ $g(f(-2))=g(-3)=(-3{)}^2=9\text{.}$ Therefore, our answer is $g(f(-2))=9\text{.}$

Suppose $f(x)=x^2-3\text{,}$ and $g(x)$ and $h(x)$ are given in the table below.

Let’s find $f(h(g(1)))\text{.}$ We start from the inside, so we begin with $g(x)\text{.}$ From the table, we can see that $g(1)=2\text{.}$ Now, we can plug that into the next function, $h(x)\text{:}$ $h(g(1))=h(2)=4\text{.}$ Finally, we plug that into the most outside function, $f(x)\text{:}$ $f(h(g(1)))=f(h(2))=f(4)=(4{)}^2-3=13\text{.}$