17Calculus Precalculus - Composite Functions
If you watched the function notation video on the main functions page, you got a taste of a concept called composite functions or composition of functions. The idea is that, instead of just plugging numbers in functions, you can also plug in more sophisticated things including other functions. That's basically all this is going on.
If you want a complete lecture on this topic, we recommend this video. He also gives you some information about operations on functions, which is good to review at this point.
Prof Leonard - Operations and Composition of Functions
video by Prof Leonard |
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Getting Started
Let's get started by looking at an example.
For the functions \(f(x)=3x^2\) and \(g(x)=x+3\), by now you know that \(f(3)=3(3)^2=27\) and \(g(2)=2+3=5\). But what is \(f(x+3)\)? Well, if you compare \(f(x+3)\) and \(f(x)\), you notice that we replaced \(x\) in \(f(x)\) with \(x+3\) to get \(f(x+3)\). So since we replace \(x\) with \(x+3\) in \(f(x)\), we do the same replacement in the function \(f(x)=3x^2\) to get \(f(x+3)=3(x+3)^2\).
Similarly, we can find \(g(3x^2)\) by replacing x with \(3x^2\) to get \(g(3x^2)=3x^2+3\).
Given the same two functions \(f(x)=3x^2\) and \(g(x)=x+3\), we can write the same two composition of functions as \(f(g(x))\) and \(g(f(x))\).
Here is a great video explaining composite functions with lots of examples.
MIP4U - Composite Functions [8min-55secs]
video by MIP4U |
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Another way to write the composition of the two functions \(f(x)\) and \(g(x)\) is \( f(g(x)) = (f \circ g)(x)\). You will also see this written as \(f \circ g\) without the \((x)\).
Be careful with composition, \(f(g(x)) \neq g(f(x))\) most of the time. There will be times that they are equal but those are very special cases (like with inverse functions).
Time for some practice.
Unless otherwise instructed, for \( f(x) = 3x - 1 \) and \( g(x) = 2x^2 + x + 1 \), find \( f \circ g \) and \( g \circ f \).
Problem Statement |
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Unless otherwise instructed, for \( f(x) = 3x - 1 \) and \( g(x) = 2x^2 + x + 1 \), find \( f \circ g \) and \( g \circ f \).
Solution |
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2916 video
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Unless otherwise instructed, for \( f(x) = 3x - 4 \) and \( g(x) = x^3 - 3 \), find \( f \circ g \) and \( g \circ f \).
Problem Statement |
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Unless otherwise instructed, for \( f(x) = 3x - 4 \) and \( g(x) = x^3 - 3 \), find \( f \circ g \) and \( g \circ f \).
Solution |
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2917 video
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Unless otherwise instructed, for \( f(x) = 5x + 2 \) and \( g(x) = x^3 - 4 \), calculate \( f(g(2)) \) and \( g(f(-1)) \).
Problem Statement |
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Unless otherwise instructed, for \( f(x) = 5x + 2 \) and \( g(x) = x^3 - 4 \), calculate \( f(g(2)) \) and \( g(f(-1)) \).
Solution |
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2918 video
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Unless otherwise instructed, for \( f(x) = 2x - 3 \) and \( g(x) = 5x + 1 \), find \( f \circ g \).
Problem Statement |
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Unless otherwise instructed, for \( f(x) = 2x - 3 \) and \( g(x) = 5x + 1 \), find \( f \circ g \).
Solution |
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2919 video
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Unless otherwise instructed, for \( f(x) = x + 3 \) and \( g(x) = x^2 - 5 \), find \( f \circ g \) and \( g \circ f \).
Problem Statement |
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Unless otherwise instructed, for \( f(x) = x + 3 \) and \( g(x) = x^2 - 5 \), find \( f \circ g \) and \( g \circ f \).
Solution |
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2920 video
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Unless otherwise instructed, for \( f(x) = 3x + 2 \) and \( g(x) = x^2 + 1 \), find \( f \circ g \) and \( g \circ f \).
Problem Statement |
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Unless otherwise instructed, for \( f(x) = 3x + 2 \) and \( g(x) = x^2 + 1 \), find \( f \circ g \) and \( g \circ f \).
Solution |
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2921 video
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Unless otherwise instructed, for \( f(x) = x^2 - 3 \) and \( g(x) = \sqrt{x-1} \), find \( f \circ g \) and \( g \circ f \).
Problem Statement |
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Unless otherwise instructed, for \( f(x) = x^2 - 3 \) and \( g(x) = \sqrt{x-1} \), find \( f \circ g \) and \( g \circ f \).
Solution |
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2922 video
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For \(f(x)=x^2+x\), \(g(x)=4-x\), find \(f \circ g\) and \(g \circ f\).
Problem Statement |
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For \(f(x)=x^2+x\), \(g(x)=4-x\), find \(f \circ g\) and \(g \circ f\).
Solution |
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1624 video
video by PatrickJMT |
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For \(f(x)=x^2-2x+3\), \(g(x)=2x+1\), find \(f \circ g\) and \(g \circ f\).
Problem Statement |
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For \(f(x)=x^2-2x+3\), \(g(x)=2x+1\), find \(f \circ g\) and \(g \circ f\).
Solution |
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1626 video
video by MIP4U |
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\(f(x)=2x-3\), \(g(x)=6x^2-4x+5\), find \(f(g(7))\), \(g(f(2))\), \(f(g(-3))\), \(f(f(-4))\).
Problem Statement |
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\(f(x)=2x-3\), \(g(x)=6x^2-4x+5\), find \(f(g(7))\), \(g(f(2))\), \(f(g(-3))\), \(f(f(-4))\).
Solution |
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1627 video
video by MIP4U |
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For \(f(x)=x^2+5\), \(g(x)=6x-15\), find \((f \circ g)(2)\) and \((g \circ f)(2)\).
Problem Statement |
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For \(f(x)=x^2+5\), \(g(x)=6x-15\), find \((f \circ g)(2)\) and \((g \circ f)(2)\).
Solution |
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1628 video
video by MIP4U |
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For \(f(x)=3x/5+4\), \(g(x)=2x^2-5x+9\), find \((f \circ g)(2)\), \((g \circ f)(4)\), \((f \circ g)(1/2)\), \((f \circ f)(-4/5)\).
Problem Statement |
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For \(f(x)=3x/5+4\), \(g(x)=2x^2-5x+9\), find \((f \circ g)(2)\), \((g \circ f)(4)\), \((f \circ g)(1/2)\), \((f \circ f)(-4/5)\).
Solution |
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1631 video
video by MIP4U |
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For \(f(x)=1/(x+5)\), \(g(x)=3/x-5\), find \(f \circ g\) and \(g \circ f\).
Problem Statement |
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For \(f(x)=1/(x+5)\), \(g(x)=3/x-5\), find \(f \circ g\) and \(g \circ f\).
Solution |
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1629 video
video by MIP4U |
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For \(f(x)=x^3-x+1\), \(g(x)=2x^2\), \(h(x)=\sqrt{x}\), find \( f(g(h(x))) \).
Problem Statement |
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For \(f(x)=x^3-x+1\), \(g(x)=2x^2\), \(h(x)=\sqrt{x}\), find \( f(g(h(x))) \).
Solution |
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1630 video
video by MIP4U |
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For \(f(x)=x/(x+1)\), \(g(x)=x^{10}\), \(h(x)=x+3\), find \(f \circ g \circ h\).
Problem Statement |
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For \(f(x)=x/(x+1)\), \(g(x)=x^{10}\), \(h(x)=x+3\), find \(f \circ g \circ h\).
Solution |
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1632 video
video by Krista King Math |
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For \(h(x)=3x\), \(g(t)=-2t-2-h(t)\), \(f(n)=-5n^2+h(n)\), find \(h(g(8))\).
Problem Statement |
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For \(h(x)=3x\), \(g(t)=-2t-2-h(t)\), \(f(n)=-5n^2+h(n)\), find \(h(g(8))\).
Solution |
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1633 video
video by Khan Academy |
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Domain of Composite Functions |
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To find the domain of composite functions, some care is required. Some instructors tell you to find the expression of the composite function and determine the domain from that. However, sometimes information is lost in the algebra and it is easy to get the incorrect answer.
The best way to determine the domain is to start with the inside function, determine the domain and range of it and then use it's range as the domain of the outside function.
Unless otherwise instructed, for \( f(x) = 1/x \) and \( g(x) = x + 3 \), find \( (f \circ g)(x) \), \( (g \circ f)(x) \) and the domains of each composite function.
Problem Statement |
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Unless otherwise instructed, for \( f(x) = 1/x \) and \( g(x) = x + 3 \), find \( (f \circ g)(x) \), \( (g \circ f)(x) \) and the domains of each composite function.
Solution |
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2923 video
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Unless otherwise instructed, for \( f(x) = \sqrt{x-2} \) and \( g(x) = 1/(x+3) \), find \( f \circ g \) and it's domain.
Problem Statement |
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Unless otherwise instructed, for \( f(x) = \sqrt{x-2} \) and \( g(x) = 1/(x+3) \), find \( f \circ g \) and it's domain.
Solution |
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2924 video
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Unless otherwise instructed, for \( f(x) = \sqrt{x-5} \) and \( g(x) = 1/(8-x) \), find \( g \circ f \) and it's domain.
Problem Statement |
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Unless otherwise instructed, for \( f(x) = \sqrt{x-5} \) and \( g(x) = 1/(8-x) \), find \( g \circ f \) and it's domain.
Solution |
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2925 video
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For \(f(x)=x+1/x\), \(g(x)=(x+1)/(x+2)\), find \(f \circ g\), \(g \circ f\), \(f \circ f\), \(g \circ g\) and their domains.
Problem Statement |
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For \(f(x)=x+1/x\), \(g(x)=(x+1)/(x+2)\), find \(f \circ g\), \(g \circ f\), \(f \circ f\), \(g \circ g\) and their domains.
Solution |
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1625 video
video by Krista King Math |
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For \(\displaystyle{f(x)=x+\frac{1}{x}; }\) \(\displaystyle{ g(x)=\frac{x+1}{x+2}}\) find the domains of \(f\circ g,~~ g\circ f,~~f\circ f,~~ g\circ g\).
Problem Statement |
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For \(\displaystyle{f(x)=x+\frac{1}{x}; }\) \(\displaystyle{ g(x)=\frac{x+1}{x+2}}\) find the domains of \(f\circ g,~~ g\circ f,~~f\circ f,~~ g\circ g\).
Solution |
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444 video
video by Krista King Math |
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Find the domain of \(g(f(x))\) where \(\displaystyle{f(x)= \frac{1}{x}; g(x)=\sqrt{x+4}}\).
Problem Statement |
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Find the domain of \(g(f(x))\) where \(\displaystyle{f(x)= \frac{1}{x}; g(x)=\sqrt{x+4}}\).
Solution |
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439 video
video by PatrickJMT |
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Find the domain of \(g(f(x))\) where \(\displaystyle{f(x)=\sqrt{x-8}; g(x)=x^2}\).
Problem Statement |
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Find the domain of \(g(f(x))\) where \(\displaystyle{f(x)=\sqrt{x-8}; g(x)=x^2}\).
Solution |
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440 video
video by PatrickJMT |
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Find the domain of \(g(f(x))\) where \(\displaystyle{f(x)=\frac{1}{x}; }\) \(\displaystyle{ g(x)=\frac{1}{(x+2)(x-3)}}\).
Problem Statement |
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Find the domain of \(g(f(x))\) where \(\displaystyle{f(x)=\frac{1}{x}; }\) \(\displaystyle{ g(x)=\frac{1}{(x+2)(x-3)}}\).
Solution |
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438 video
video by PatrickJMT |
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Really UNDERSTAND Precalculus
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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