Gib für folgende Funktionen die maximale Definitionsmenge an (G=R)\left(G=ℝ\right)(G=R) .
f(x)=5x+170,64x2+1,12x+0,49f\left(x\right)=\dfrac{5x+17}{0{,}64x^2+1{,}12x+0{,}49}f(x)=0,64x2+1,12x+0,495x+17
f(x)=x2+4x+42x2+20x+60f\left(x\right)=\dfrac{\sqrt{x^2+4x+4}}{\sqrt{2x^2+20x+60}}f(x)=2x2+20x+60x2+4x+4
f(x)=(4x−9)1218−8xf\left(x\right)=\dfrac{\left(4x-9\right)^\tfrac12}{18-8x}f(x)=18−8x(4x−9)21
f(x)=6x−x2−9f\left(x\right)=\sqrt{6x-x^2-9}f(x)=6x−x2−9
f(x)=112x2−2f\left(x\right)=\dfrac1{\frac12x^2-2}f(x)=21x2−21
f(x)=x3+x2+x+149x2−14f\left(x\right)=\dfrac{x^3+x^2+x+1}{\frac49x^2-\frac14}f(x)=94x2−41x3+x2+x+1
f(x)=sin(x)x2−4x+4f\left(x\right)=\dfrac{\sin(x)}{x^2-4x+4}f(x)=x2−4x+4sin(x)
f(x)=1−x2+6x−9f(x)=\dfrac1{-x^2+6x-9}f(x)=−x2+6x−91
f(x)=2ax+4bx2+8cx380−5x2f\left(x\right)=\dfrac{2\mathrm{ax}+4\mathrm{bx}^2+8\mathrm{cx}^3}{80-5x^2}f(x)=80−5x22ax+4bx2+8cx3
f(x)=(2−x)⋅x+x2+x3+x4x3−14xf\left(x\right)=\left(2-x\right)\cdot\dfrac{x+x^2+x^3+x^4}{x^3-\frac14x}f(x)=(2−x)⋅x3−41xx+x2+x3+x4
f(x)=xsin(x)f\left(x\right)=\dfrac x{\sin(x)}f(x)=sin(x)x
f(x)=123⋅4+5x+6x2cos(x+4)f\left(x\right)=123\cdot\dfrac{4+5x+6x^2}{\cos\left(x+4\right)}f(x)=123⋅cos(x+4)4+5x+6x2
f(x)=67893 cos(x)−sin(x)f\left(x\right)=\dfrac{6789}{\sqrt3\;\cos\left(x\right)-\sin\left(x\right)}f(x)=3cos(x)−sin(x)6789
f(x)=5x2−a36x2−16xf(x)=\dfrac{5x^2-a}{36x^2-16x}f(x)=36x2−16x5x2−a
f(x)=1x−6−16x+1f(x)=\dfrac1{x-6}-\dfrac1{6x+1}f(x)=x−61−6x+11
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