1105 RP2 084 – Evaluating a Composition of Functions from Expressions
Articles Blog

1105 RP2 084 – Evaluating a Composition of Functions from Expressions

September 25, 2019


– [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.

Leave a Reply

Your email address will not be published. Required fields are marked *