Rozwiązanie:
[tex]d)[/tex]
[tex]$f(x)=\frac{log(x^{3}-3x+2)}{2^{x}}[/tex]
Dziedzina:
[tex]x^{3}-3x+2>0 \wedge 2^{x}-4\neq 0[/tex]
Pierwszy warunek:
[tex]x^{3}-3x+2>0\\x^{3}-x-2x+2>0\\x(x^{2}-1)-2(x-1)>0\\x(x-1)(x+1)-2(x-1)>0\\(x-1)(x(x+1)-2)>0\\(x-1)(x^{2}+x-2)>0\\(x-1)^{2}(x+2)>0\\x \in (-2,1) \cup (1,\infty)[/tex]
Drugi warunek:
[tex]2^{x}\neq 4\\2^{x}\neq 2^{2} \iff x\neq 2[/tex]
Zatem:
[tex]D=\{x \in \mathbb{R}:x \in\ (-2,1) \cup (1,2) \cup (2,\infty)\}}[/tex]
[tex]e)[/tex]
[tex]$f(x)=\sqrt{\frac{x-1}{x} } +\frac{3x}{lnx}[/tex]
Dziedzina:
[tex]$\frac{x-1}{x} \geq 0 \ \wedge \ x>0 \ \wedge \ x\neq 0 \ \wedge \ lnx\neq 0[/tex]
Pierwszy warunek:
[tex]$\frac{x-1}{x} \geq 0 \iff x(x-1)\geq 0[/tex]
[tex]x \in (-\infty,0 \rangle \ \cup \ \langle 1,\infty)[/tex]
Czwarty warunek:
[tex]lnx\neq 0\\x\neq 1[/tex]
Zatem:
[tex]D=\{x \in \mathbb{R}:x \in (1,\infty)\}[/tex]
[tex]g)[/tex]
[tex]g(x)=\sqrt{ln^{2}x-lnx-2}[/tex]
Dziedzina:
[tex]ln^{2}x-lnx-2>0 \wedge x>0[/tex]
Pierwszy warunek:
Niech [tex]t=lnx[/tex] :
[tex]t^{2}-t-2\geq 0\\(t+1)(t-2)\geq 0\\t \in (-\infty,-1 \rangle \ \cup \langle 2,\infty)[/tex]
Stąd:
[tex]lnx\leq -1 \wedge lnx\geq 2\\lnx\leq ln(e^{-1}) \wedge lnx\geq ln(e^{2})\\x\leq e^{-1} \wedge x\geq e^{2}\\x \in \ (-\infty,e^{-1} \rangle \ \cup \ \langle e^{2},\infty)[/tex]
Zatem:
[tex]$D=\{x \in \mathbb{R}:x \in (0,\frac{1}{e} \rangle \cup \langle e^{2},\infty) \}[/tex]
[tex]h)[/tex]
[tex]$f(x)=\sqrt{sin2x}[/tex]
Dziedzina:
[tex]sin2x\geq 0\\[/tex]
[tex]$x \in \langle k\pi ,\frac{\pi}{2}+k\pi \rangle \ \ k \in \mathbb{Z}[/tex]
[tex]i)[/tex]
[tex]$f(x)=\frac{1}{\sqrt{1-2sinx} } +\sqrt{\frac{3\pi }{2} - |x-\frac{\pi }{2} |}[/tex]
Dziedzina:
[tex]$1-2sinx>0 \wedge \frac{3\pi }{2} - |x-\frac{\pi }{2} |\geq 0[/tex]
Pierwszy warunek:
[tex]1-2sinx>0[/tex]
[tex]$sinx<\frac{1}{2}[/tex]
[tex]$x \in ( -\frac{7\pi }{6} +2k\pi ,\frac{\pi }{6} +2k\pi ) \ \ k \in \mathbb{Z}[/tex]
Drugi warunek:
[tex]$ \frac{3\pi }{2} - |x-\frac{\pi }{2} |\geq 0[/tex]
[tex]$ |x-\frac{\pi }{2} |\leq \frac{3\pi }{2}[/tex]
[tex]$x-\frac{\pi }{2} \leq \frac{3\pi }{2} \wedge x-\frac{\pi }{2} \geq -\frac{3\pi }{2}[/tex]
[tex]x\leq 2\pi \wedge x\geq -\pi[/tex]
[tex]x \in \langle -\pi ,2\pi \rangle[/tex]
Zatem:
[tex]$D=\{x \in \mathbb{R}: x \in \langle -\pi ,\frac{\pi }{6}) \cup (\frac{5\pi}{6},2\pi \rangle \}[/tex]
