ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: f0od on February 17, 2019, 08:24:57 pm
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I don't really understand the fogs & gofs of composite functions – in particular finding the range of a composite function
If anyone would be able to help with the questions below it would be much appreciated!
1) Given the functions f(x) = 1/x and g(x) = x^2 + 1, find the (domain) and range of f(g(x))
- So far I've gotten that the domain of this composite function is R/all real numbers
2) Given the functions f(x) = sin(x), x∈[0, 2pi] and g(x) = x^2 - 4, x∈[-2,2], find the domain and range of g(f(x))
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I don't really understand the fogs & gofs of composite functions – in particular finding the range of a composite function
If anyone would be able to help with the questions below it would be much appreciated!
1) Given the functions f(x) = 1/x and g(x) = x^2 + 1, find the (domain) and range of f(g(x))
- So far I've gotten that the domain of this composite function is R/all real numbers
2) Given the functions f(x) = sin(x), x∈[0, 2pi] and g(x) = x^2 - 4, x∈[-2,2], find the domain and range of g(f(x))
To find the range of a composite function f(g(x)):
1. Find the implied domain of f(g(x)). Call this set X.
2. Find the range of g(x) over X. Call this set Y.
3. Find the range of f(x) over Y. This gives the range of f(g(x)).
For example, let f(g(x)) = 1/(x^2+1).
1. The implied domain of f(g(x)) is R
2. The range of x^2 +.1 over R is [1, infinity)
3. The range of 1/x over [1, infinity) is (0, 1].
Hence, the range of 1/(x^2+1) over its implied domain is (0, 1].
A similar method works with your second example, sin^2(x) – 4.
1. The domain is [0, 2pi].
2. The range of sin(x) over R is [–1, 1]. (Notice this is a subset of [–2, 2], the domain of g(x) = x^2 – 4. If not, we'd have to restrict the domain of sin(x) so that its range is a subset of [–2, 2]. See 2017 Exam 1 Q7 for a question along these lines).
3. The range of x^2 – 4 over [–1, 1] is [–4, –3].
Hence, the range of sin^2(x) – 4 over [0, 2pi] is [–4, –3].
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I don't really understand the fogs & gofs of composite functions – in particular finding the range of a composite function
If anyone would be able to help with the questions below it would be much appreciated!
1) Given the functions f(x) = 1/x and g(x) = x^2 + 1, find the (domain) and range of f(g(x))
- So far I've gotten that the domain of this composite function is R/all real numbers
2) Given the functions f(x) = sin(x), x∈[0, 2pi] and g(x) = x^2 - 4, x∈[-2,2], find the domain and range of g(f(x))
Question 1
\[f:\mathbb{R}\setminus\{0\}\to\mathbb{R},\ f(x)=\frac{1}{x}\quad\text{and}\quad g:\mathbb{R}\to\mathbb{R},\ g(x)=x^2+1\]Here, \(\text{range}(g)=[1,\infty)\subseteq \text{domain}(f)=\mathbb{R}\setminus\{0\}\), so \(f(g(x))\) is defined.
As you already said, \(\text{domain}(f\circ g)=\text{domain}(g)=\mathbb{R}\).
Let us just take a step back for a second and think about what we're actually doing here.
We plug in a real number \(x\) into \(g\) and the output is somewhere in \([1,\infty)\). This output of \(g\) then goes in as the input for \(f\).
Essentially: \(\mathbb{R}\overset{g}{\longrightarrow}[1,\infty)\overset{f}{\longrightarrow}\text{range}(f\circ g)\).
In order words, \(\text{range}(f\circ g)=\text{range}(f^*)\), where \[f^*:[1,\infty)\to\mathbb{R},\ f^*(x)=\frac{1}{x}.\] A quick sketch of the graph of \(f^*\) shows that \(\text{range}(f\circ g)=(0,1]\).
Question 2
This is the same sort of idea as Question 1. I'll start you off, and I'll leave it with you. \[f:[0,2\pi]\to\mathbb{R},\ f(x)=\sin(x)\quad \text{and}\quad g:[-2,2]\to\mathbb{R},\ g(x)=x^2-4.\] Clearly, \(\text{range}(f)=[-1,1]\subseteq \text{domain}(g)=[-2,2]\), so \(g(f(x))\) is defined.
As before, \(\text{domain}(g\circ f)=\text{domain}(f)=[0,2\pi]\).
In this case: \([0,2\pi]\overset{f}{\longrightarrow}[-1,1]\overset{g}{\longrightarrow}\text{range}(g\circ f)\).
So, what's the range of \(g\circ f\) here?
Edit: S_R_K by milliseconds lol
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I feel like i should delete my post in light of Dan's beautifully presented explanation.