Aufgaben zur Bestimmung von Definitionsmengen

$f(x)=\dfrac{4x-3}{2x-5}$

$f(x)=\dfrac{3x^3+7}{x^2-2x}$

$f\left(x\right)=\dfrac{2x+x^7}{\frac19x^2+\frac13x+\frac14}$

$f\left(x\right)=\dfrac{2x^4-x^2}{0{,}01x^2-1}$

$f\left(x\right)=\sqrt{7x+4}$

$f\left(x\right)=\sqrt{x^2-5x+6}$

$f(x)=\sqrt{17x}+5x-3$

$f(x)=\sqrt[6]{x^2-4x+3}$

$f(x)=\ln(x-5)$

$f\left(x\right)=\ln\left(6x-x^2-9\right)$

$f(x)=\mathrm{log}_6(x^3-7x)$

$f\left(x\right)=5x\;\tan\left(x\right)$

$f\left(x\right)=7x^2\tan\left(2-x\right)$

$f\left(x\right)=\left(x+5\right)\;\tan\;\left(x^2-\frac12\pi\right)$

$f\left(x\right)=\dfrac1{\sqrt{x^2+6x+9}}$

$f\left(x\right)=\dfrac{\lg\left(x^2-x\right)}{\sqrt{x+2}}$

$f\left(x\right)=\dfrac{1+3x+33x^2}{\sin\left(x\right)}$

$f\left(x\right)=\dfrac{12345}{1-\sin\left(x-\frac12\pi\right)}$

$f\left(x\right)=\dfrac{5x+17}{0{,}64x^2+1{,}12x+0{,}49}$

$f\left(x\right)=\dfrac{\sqrt{x^2+4x+4}}{\sqrt{2x^2+20x+60}}$

$f\left(x\right)=\dfrac{\left(4x-9\right)^\tfrac12}{18-8x}$

$f\left(x\right)=\sqrt{6x-x^2-9}$

$f\left(x\right)=\dfrac1{\frac12x^2-2}$

$f\left(x\right)=\dfrac{x^3+x^2+x+1}{\frac49x^2-\frac14}$

$f\left(x\right)=\dfrac{\sin(x)}{x^2-4x+4}$

$f(x)=\dfrac1{-x^2+6x-9}$

$f\left(x\right)=\dfrac{2\mathrm{ax}+4\mathrm{bx}^2+8\mathrm{cx}^3}{80-5x^2}$

$f\left(x\right)=\left(2-x\right)\cdot\dfrac{x+x^2+x^3+x^4}{x^3-\frac14x}$

$f\left(x\right)=\dfrac x{\sin(x)}$

$f\left(x\right)=123\cdot\dfrac{4+5x+6x^2}{\cos\left(x+4\right)}$

$f\left(x\right)=\dfrac{6789}{\sqrt3\;\cos\left(x\right)-\sin\left(x\right)}$

$f(x)=\dfrac{5x^2-a}{36x^2-16x}$

$f(x)=\dfrac1{x-6}-\dfrac1{6x+1}$