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In $11-18 :$ a. Find $h(x)$ when $h(x)=g(f(x)) .$ b. What is the domain of $h(x) ?$ c. What is the range of $\mathrm{h}(x) ?$ d. Graph $\mathrm{h}(x)$$$\mathrm{f}(x)=5-x, \mathrm{g}(x)=|x|$$

a) $h(x)=|5-x|$b) $\mathbb{R}$c) $\mathbb{R}^{+}$d) See graph

Algebra

Chapter 4

RELATIONS AND FUNCTIONS

Section 7

Composition of Functions

An Introduction to Geometry

Functions

Linear Functions

Polynomials

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So for this exercise, we need to find a chav X to find his G of f of X, where FX is equal to the absolute value of two plus X and G of X is equal to negative X. So I'm gonna start by plugging F of X into G of x o g f of X equals negative. And then what goes in here? The two plus X, perhaps a value to plus sex. So we have to find the domain and there's no restriction on any of the X values we can choose. The domain is all rials, and we'll notice for the range, which is all the possible output values, that there's no X value we can pick that will make this function evaluate to a positive number. If we were to pick negative to, that would be two plus negative too, which is zero. If we were to pick, uh, negative three. We have to close a negative three, which is negative. One absolute value of negative one is one, but then negated. It's negative one again. So our range is actually all why values less than or equal to zero and the graph of this function looks something like this. You'll see if we plug in negative too. We get zero. If we plug in zero, we get negative too. And if we plug in negative four, we get to minus negative for negative to absolute value of negative to positive. To negated is also negative too. So that's the graph of dysfunction right here, which we've defined as H of X.

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