Tema 3. Funcions contínues
En aquest tema tractarem les funcions exponencials, logarítmiques, trigonomètriques i hiperbòliques.
Funcions exponencials
[math]\displaystyle{ a\gt 0, a \in \mathbb{R} }[/math]
[math]\displaystyle{ n \in \mathbb{Z}, a^n }[/math]
[math]\displaystyle{ n \in \mathbb{Z}, a^n \quad a^{-1}=\frac{1}{a} }[/math]
[math]\displaystyle{ \frac{p}{q}=n \in \mathbb{Q}, a^n = ? \rightarrow x = \sqrt[q]{a^p} \Rightarrow x^q = a^p }[/math]
[math]\displaystyle{ r \in \mathbb{R}, a^r = ? }[/math]
Teorema:
[math]\displaystyle{ \forall a \in \mathbb{R}, a\gt 0, }[/math] existeix una única funció contínua definida a [math]\displaystyle{ \mathbb{R} }[/math] [math]\displaystyle{ (f:\mathbb{R} \rightarrow \mathbb{R}) }[/math] tal que:
- [math]\displaystyle{ f(x+y)=f(x)\cdot f(y) \quad (a^{x+y}=a^x \cdot a^y) }[/math]
- [math]\displaystyle{ f(1)=a }[/math]
Es pot comprovar que [math]\displaystyle{ f(\frac{p}{q}) = a^\frac{p}{q} \forall \frac{p}{q} \in \mathbb{Q} }[/math]. Llavors escrivim [math]\displaystyle{ f(x) = a^x \forall x \in \mathbb{R} }[/math] i s'anomena funció exponencial de base "a".
Propietats
- [math]\displaystyle{ a^0 = 1; a^1 = a }[/math]
- [math]\displaystyle{ a^{x+y}=a^x \cdot a^y }[/math]
- [math]\displaystyle{ (ab)^x = a^x \cdot b^x }[/math]
- [math]\displaystyle{ (a^x)^y = a^{x \cdot y} }[/math]
- [math]\displaystyle{ 1^x = 1 \forall x }[/math]
Gràfica de [math]\displaystyle{ a^x }[/math]
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Obs: [math]\displaystyle{ a^x }[/math] és una funció bijectiva (si [math]\displaystyle{ a \neq 1 }[/math]) i [math]\displaystyle{ \begin{array}{rl} f:&\mathbb{R} \longrightarrow (0, +\infty) \\ & x \longmapsto a^x=y\end{array} \implies \text{ Té inversa: } \begin{array}{rl} g: & (0, +\infty) \longrightarrow \mathbb{R} \\ & y \longmapsto x \end{array} }[/math]
La funció inversa a l'exponencial s'anomena logaritme en base "a".
Funcions logarítmiques
[math]\displaystyle{ \log_a: (0, +\infty) \longrightarrow \mathbb{R} }[/math]
Propietats
- [math]\displaystyle{ \log_a 1 = 0; \log_a a = 1 }[/math]
- [math]\displaystyle{ \log_a (xy) = \log_a(x) + \log_a(y) }[/math]
- [math]\displaystyle{ \log_a (x^y) = y \cdot \log_a x }[/math]
