– [Narrator] Given F of X

is equal to two X plus one, and G of X is equal to the

polynomial X minus one, over the polynomial X plus

two, find A, F of G of one. Our first step will be to compute G of one using the formula for G. We get the expression one minus one, over the expression one plus two. That simplifies to zero

over three, which is zero. We now substitute zero, for G of one. Zero was an output for G,

but now it’s an input for F, so, we substitute zero

into the formula for F. We get two times zero plus

one, which simplifies to one. Therefore, F of G of one is equal to one. For B, we’re asked to find

G of F of negative two. So, first of all, we

find F of negative two. We get two times negative

two, which is negative four, plus one, which simplifies

to negative three. So, we substitute negative

three for F of negative two. Now, we need to find G of negative three. We get the expression

negative three minus one, all over the expression

negative three plus two. That simplifies to negative

four over negative one, which is positive four. So, G of F of negative

two is equal to four. For C, we’re asked to find F of G of two. We first find G of two. We get the expression two minus one over the expression two plus two. That simplifies to one

over four, which is 1/4. So, we substitute 1/4 for G of two. Now, we need to find F of 1/4. We get two times 1/4 plus one. That simplifies to 1/2 plus one, which is one and a half, or three halves. So, F of G of two is

equal to three halves. For part D, we’re asked

to find F of F of zero. So, first of all, we find F of zero. We get two times zero plus

one, which simplifies to one. We now substitute one for F of zero. One was the output of F,

but now it’s the input of F, so we need to evaluate F of one. We get two times one plus one,

which simplifies to three. So, F of F of zero is equal to three. For part E, we’re asked to

evaluate F circle G of X. Well, what we need to recall

is that F circle G of X is the composition of F and G. We need to find F of G of X. The output for G of X is X minus one over the expression X plus two. We’re gonna take that output and substitute it as the input for F. We get two times the expression, X minus one, all over

X plus two, plus one. To combine these two terms, we get a common denominator of X plus two. For the first term, we distribute two through the parentheses, and we get two times X minus one is equal to two X minus

two, and one can be written as X plus two over X plus two. Combining the two rational

expressions together, we get three X all over X plus two. For part F, we need to

find G circle F of X, we need to find the

composition of G and F, so we need to evaluate G of F of X. F of X represents the output

of F when the input is X. That’s given by the

expression two X plus one. We’re gonna take that output

and make it the input for G. We get the expression

two X plus one minus one, over the expression two

X plus one plus two. The numerator simplifies to two X. The denominator simplifies

to two X plus three. So, the final result is two

X over two X plus three.