New Functions from Old

Composing Functions

Sum, Difference, Products and Quotients

$f:A\to B$ and $g:C\to D$

• Sum, Difference, Products
  • Domain: $A\cap C$
• Quotients
  • Domain: $(A\cap C)\setminus \{x\in C:g(x)=0\}$

$f(x)=|ln(x)|$ … Range: ${\mathbb{R}}_0^{+}$ … Domain ${\mathbb{R}}^{+}$ $g(x)=sin(x)$ … Range: $[-1,1]$ … Domain $\mathbb{R}$

• $(f+g)(x)$ … Range: $\subseteq [-1,\infty )$
• $(f-g)(x)$ … Range: $\subseteq [-1,\infty )$
• $(f\ast g)(x)$ … Range: $\subseteq \mathbb{R}$
• $(f/g)(x)$ … Range: $\subseteq \mathbb{R}$ … Domain: ${\mathbb{R}}^{+}\setminus \{k\ast \pi ,k\in \mathbb{N}\}$
  • $\{k\ast \pi ,k\in \mathbb{N}\}$ because everywhere, where $k$ is a multiple of $\pi$, the result of $sin(k)$ will be $0$. Therefore we have to exclude them

Composition

$f\circ g\Rightarrow f(g(x))$ … plug x into $g$ first, then plug the result into $f$.

$f$ … exterior function $g$ … interior function

$f\circ g\ne f\ast g$ … also $f^2\ne f\circ f$ $f\circ g\ne g\circ f$ they might be the same, but it is not generally true

$f(x)=3x-x^3$ and $g(x)=x^3$

$(f\circ g)(x)=f(g(x))=2$ $(g\circ f)(x)=g(f(x))=8$ therefore … $f\circ g\ne g\circ f$

Injections, Surjections, Bijections

• A function $f$ is called injective if $\forall x_1\ne x_2\in A\Rightarrow f(x_1)\ne f(x_2)$ - this needs to hold for any $x_1$ and $x_2$ combination. - A parabola $f(x)=x^2$ is not injective, because at $x=1$ and $x=-1$ the result is the same ($1$)
  • One can check by drawing horizontal lines and checking, if any line crosses the function at least twice
  • typical non-injective functions: $sin,cos,x^2$
  • depends also on Domain, $x^2$ may be injective if only positive $x$ values are allowed ${\mathbb{R}}^{+}$
  • Injectivity
• A function is called surjective if its range is equal to its codomain
• A function is called bijective if it is both injective and surjective

Identity Function

$I{d}_A=f(x)=x$ … nothing happens

Inverse Functions

$f:A\to B$ and $g:B\to A$

$g$ is called inverse if $g\circ f=I{d}_A$ and $f\circ d=I{d}_B$

notation: $f=f^{-1}$ careful: $f^{-1}\ne \frac{1}{f}$ (inverse != )

graphically find the inverse by reflecting the function $f$ around the Identity Function of the Domain of $f$.

$f(x)=x^2-1=y$ … solve for x $x^2=y+1$ $x=\sqrt{y+1}$ $f^{-1}(y)=\sqrt{y+1}$ … done

another example $f(x)=3ln(\sqrt{x+4}-2)$ Domain: $(0,\infty )$ Range: $\mathbb{R}$ after solving … $f^{-1}=(exp(\frac{x}{4})+2{)}^2-4$ Domain: $(0,\infty )$ … $(exp(\frac{x}{4})+2{)}^2$ will always be slightly larger than $4$ Range: $\mathbb{R}$