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Using [link] , evaluate $\text{\hspace{0.17em}}g(f(2)).$
$g(f(2))=g(5)=3$
When evaluating a composite function where we have either created or been given formulas, the rule of working from the inside out remains the same. The input value to the outer function will be the output of the inner function, which may be a numerical value, a variable name, or a more complicated expression.
While we can compose the functions for each individual input value, it is sometimes helpful to find a single formula that will calculate the result of a composition $\text{\hspace{0.17em}}f\left(g\left(x\right)\right).\text{\hspace{0.17em}}$ To do this, we will extend our idea of function evaluation. Recall that, when we evaluate a function like $\text{\hspace{0.17em}}f(t)={t}^{2}-t,\text{\hspace{0.17em}}$ we substitute the value inside the parentheses into the formula wherever we see the input variable.
Given a formula for a composite function, evaluate the function.
Given $\text{\hspace{0.17em}}f(t)={t}^{2}-t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h(x)=3x+2,\text{\hspace{0.17em}}$ evaluate $\text{\hspace{0.17em}}f(h(1)).$
Because the inside expression is $\text{\hspace{0.17em}}h(1),\text{\hspace{0.17em}}$ we start by evaluating $\text{\hspace{0.17em}}h(x)\text{\hspace{0.17em}}$ at 1.
Then $\text{\hspace{0.17em}}f(h(1))=f(5),\text{\hspace{0.17em}}$ so we evaluate $\text{\hspace{0.17em}}f(t)\text{\hspace{0.17em}}$ at an input of 5.
Given $\text{\hspace{0.17em}}f(t)={t}^{2}-t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h(x)=3x+2,\text{\hspace{0.17em}}$ evaluate
a. 8; b. 20
As we discussed previously, the domain of a composite function such as $\text{\hspace{0.17em}}f\circ g\text{\hspace{0.17em}}$ is dependent on the domain of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ and the domain of $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$ It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as $\text{\hspace{0.17em}}f\circ g.\text{\hspace{0.17em}}$ Let us assume we know the domains of the functions $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ separately. If we write the composite function for an input $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}f\left(g\left(x\right)\right),\text{\hspace{0.17em}}$ we can see right away that $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ must be a member of the domain of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ must be a member of the domain of $\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}$ otherwise the second function evaluation in $\text{\hspace{0.17em}}f\left(g\left(x\right)\right)\text{\hspace{0.17em}}$ cannot be completed, and the expression is still undefined. Thus the domain of $\text{\hspace{0.17em}}f\circ g\text{\hspace{0.17em}}$ consists of only those inputs in the domain of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ that produce outputs from $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ belonging to the domain of $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$ Note that the domain of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ composed with $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ is the set of all $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in the domain of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ is in the domain of $\text{\hspace{0.17em}}f.$
The domain of a composite function $\text{\hspace{0.17em}}f\left(g\left(x\right)\right)\text{\hspace{0.17em}}$ is the set of those inputs $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in the domain of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ for which $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ is in the domain of $\text{\hspace{0.17em}}f.$
Given a function composition $\text{\hspace{0.17em}}f(g(x)),$ determine its domain.
Find the domain of
The domain of $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ consists of all real numbers except $\text{\hspace{0.17em}}x=\frac{2}{3},\text{\hspace{0.17em}}$ since that input value would cause us to divide by 0. Likewise, the domain of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ consists of all real numbers except 1. So we need to exclude from the domain of $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ that value of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ for which $\text{\hspace{0.17em}}g\left(x\right)=1.$
So the domain of $\text{\hspace{0.17em}}f\circ g\text{\hspace{0.17em}}$ is the set of all real numbers except $\text{\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}2.\text{\hspace{0.17em}}$ This means that
We can write this in interval notation as
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