[TOC] # 2.3 - Composition of functions - Given $f(x)$ and $g(x)$, we invent a new function $f \circ g$: $$(f \circ g)(x)=f(g(x))$$ - Easy way to remember: Read it from right to left. Plug $g$ into $f$. ## Composition ### Example. $f(x)=2x+1$, $g(x)=x^2$ ### 1. \begin{align} (f \circ g)(1) &=f(g(1)) \\ &=f(1^2) \\ &=f(1) \\ &=2(1)+1 \\ &=3 \end{align} ### 2. \begin{align} (f \circ g)(2) &=f(g(2)) \\ &=f(2^2) \\ &=f(4) \\ &=2(4)+1 \\ &=9 \end{align} ### 3. \begin{align} f\left(g\left(\frac{1}{2}\right)\right)=&=f\left(\left(\dfrac{1}{2}\right)^2\right) \\ &=f\left(\dfrac{1}{4}\right) \\ &=2 \left(\frac{1}{4}\right)+1 \\ &=\dfrac{2}{4}+1 \\ &=\dfrac{1}{2}+1 \\ &=\dfrac{1}{2}+\dfrac{2}{2} \\ &=\dfrac{3}{2} \end{align} ### 4. \begin{align} (f \circ g)(x)&=f(g(x)) \\ &=f(x^2) \\ &=2(x^2)+1 \\ &=2x^2+1 \end{align} - We can also do $g\circ f$: $(g\circ f)(x)=g(f(x))$ ### 5. \begin{align} (g\circ f)(2)&=g(f(2)) \\ &=g(2(2)+1) \\ &=g(5) \\ &=5^2 \\ &=25 \end{align} ### 6. \begin{align} (g \circ f)(-1) &= g(f(-1)) \\ &= g(2(-1) + 1) \\ &= g(-2 + 1) \\ &= g(-1) \\ &= (-1)^2 \\ &= 1 \end{align} ### 7. \begin{align} (g \circ f)\left(\frac{1}{2}\right) &= g\left(f\left(\frac{1}{2}\right)\right) \\ &= g\left(2\left(\frac{1}{2}\right) + 1\right) \\ &= g\left(1 + 1\right) \\ &= g(2) \\ &= 2^2 \\ &= 4 \end{align} ### 8. \begin{align} (g \circ f)(x) &= g(f(x)) \\ &= g(2x + 1) \\ &= (2x + 1)^2 \\ &= 4x^2 + 4x + 1 \end{align} - We can also do $f\circ f$ and $g\circ g$: ### 9. \begin{align} (g \circ g)(-2) &= g(g(-2)) \\ &= g((-2)^2) \\ &= g(4) \\ &= 4^2 \\ &= 16 \end{align} ### 10. \begin{align} (g \circ g)\left(\frac{1}{2}\right) &= g\left(g\left(\frac{1}{2}\right)\right) \\ &= g\left(\left(\frac{1}{2}\right)^2\right) \\ &= g\left(\frac{1}{4}\right) \\ &= \left(\frac{1}{4}\right)^2 \\ &= \frac{1}{16} \end{align} ### 11. \begin{align} (g \circ g)(x) &= g(g(x)) \\ &= g(x^2) \\ &= (x^2)^2 \\ &= x^4 \end{align} ### 12. \begin{align} (f \circ f)(-1) &= f(f(-1)) \\ &= f(2(-1) + 1) \\ &= f(-2 + 1) \\ &= f(-1) \\ &= 2(-1) + 1 \\ &= -2 + 1 \\ &= -1 \end{align} ### 13. \begin{align} (f \circ f)\left(\frac{1}{2}\right) &= f\left(f\left(\frac{1}{2}\right)\right) \\ &= f\left(2\left(\frac{1}{2}\right) + 1\right) \\ &= f(1 + 1) \\ &= f(2) \\ &= 2(2) + 1 \\ &= 4 + 1 \\ &= 5 \end{align} ### 14. \begin{align} (f \circ f)(x) &= f(f(x)) \\ &= f(2x + 1) \\ &= 2(2x + 1) + 1 \\ &= 4x + 2 + 1 \\ &= 4x + 3 \end{align} --- ## Let $f(x)=3x+1$; $g(x)=x^7$; $h(x)=\sqrt{x}$; $k(x)=\dfrac{1}{x}$. ### 15. \begin{align} (f \circ g)(x) &= f(g(x)) \\ &= f(x^7) \\ &= 3(x^7) + 1 \\ &= 3x^7 + 1 \end{align} ### 16. \begin{align} (g \circ f)(x) &= g(f(x)) \\ &= g(3x + 1) \\ &= (3x + 1)^7 \end{align} ### 17. \begin{align} (f \circ h)(x) &= f(h(x)) \\ &= f(\sqrt{x}) \\ &= 3\sqrt{x} + 1 \end{align} ### 18. \begin{align} (h \circ f)(x) &= h(f(x)) \\ &= h(3x + 1) \\ &= \sqrt{3x + 1} \end{align} ### 19. \begin{align} (f \circ k)(x) &= f(k(x)) \\ &= f\left(\frac{1}{x}\right) \\ &= 3\left(\frac{1}{x}\right) + 1 \\ &= \frac{3}{x} + 1 \end{align} ### 20. \begin{align} (k \circ f)(x) &= k(f(x)) \\ &= k(3x + 1) \\ &= \frac{1}{3x + 1} \end{align} --- ## Decomposition of Functions - Consider the way the above compositions look. In the future (for people who take calculus), we will see forms like this and be asked - what composition was this? --- ### 21. $(f\circ g)(x)=(3x-8)^5$. What are $f(x)$ and $g(x)$? |Outside Function| Inside Function| |---|---| |$f(u)=u^5$| $g(x)=3x-8$| $g$ is the inside function and $f$ is the outside function. So $g(x)=3x-8$ and $f(x)=x^5$. --- ### 22. $(f\circ g)(x)=\sqrt{2x-9}$. What are $f(x)$ and $g(x)$? |Outside Function| Inside Function| |---|---| |$f(u)=\sqrt{u}$| $g(x)=2x-9$| $g$ is the inside function and $f$ is the outside function. So $g(x)=2x-9$ and $f(x)=\sqrt{x}$. --- ### 23. $(f\circ g)(x)=\frac{1}{10x+4}$. What are $f(x)$ and $g(x)$? $g$ is the inside function and $f$ is the outside function. So $g(x)=10x+4$ and $f(x)=\dfrac{1}{x}$. |Outside Function| Inside Function| |---|---| |$f(u)=\dfrac{1}{u}$| $g(x)=10x+4$|
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