Złożenia funkcji
: 16 paź 2017, o 23:30
a)
\(\displaystyle{ f(x)= x^{2}-1 \\
Df=\RR \\
Zf=(-1, \infty )\\
g(x)=(2x+1) \\
Dg=\RR \\
Zg=\RR \\
f\circ g=f(g(x))=f(2x+1)= (2x+1)^{2}-1 \\
g\circ f=g(f(x))=g(x^{2}-1)=2(x^{2}-1)+1}\)
czy powinnam to dalej wyliczać?
b)
\(\displaystyle{ f(x)=2x+1 \\
Df=\RR \\
Zf=\RR \\
g(x)=\left| \left| x\right| \right|-1\\
Dg=\RR\setminus\{-1;1\} \\
Zg=\RR\setminus\{-1;1\}\\
f(g(x))=f(\left| \left| x\right| \right|-1)=2(\left| \left| x\right| \right|-1)+1 \\
g(f(x))=g(2x+1)=\left| \left| 2x+1\right| \right|-1}\)
c)
\(\displaystyle{ f(x)= \sqrt{x} \\
Df=\NN \\
g(x)= x^{2} \\
Dg=\RR}\)
brak złożenia?
d)
\(\displaystyle{ f(x)=\cos x \\
Df=\RR \\
g(x)=x+1 \\
Dg=\RR\setminus\{-1\} \\
f\circ g=f(g(x))=f(x+1)=\cos (x+1) \\
g\circ f=g(f(x))=g(\cos x)=(\cos x)+1}\)
e)
\(\displaystyle{ f(x)=(x-1) ^{5} \\
Df=\RR \\
g(x)= x^{2}+3x-2 \\
f\circ g=f(g(x))=f(x^{2}+3x-2)=((x^{2}+3x-2)-1) ^{5} \\
g\circ f=g(f(x))=g((x-1) ^{5})=((x-1) ^{5})^{2}+3((x-1) ^{5})-2}\)
f)
\(\displaystyle{ f(x)= x^{2}-1 \\
g(x)= \sqrt{x-1} \\
f\circ g=f(g(x))=f(\sqrt{x-1})=(\sqrt{x-1})^{2}-1 \\
g\circ f=g(f(x))=g(x^{2}-1)=\sqrt{(x^{2}-1)-1}}\)
\(\displaystyle{ f(x)= x^{2}-1 \\
Df=\RR \\
Zf=(-1, \infty )\\
g(x)=(2x+1) \\
Dg=\RR \\
Zg=\RR \\
f\circ g=f(g(x))=f(2x+1)= (2x+1)^{2}-1 \\
g\circ f=g(f(x))=g(x^{2}-1)=2(x^{2}-1)+1}\)
czy powinnam to dalej wyliczać?
b)
\(\displaystyle{ f(x)=2x+1 \\
Df=\RR \\
Zf=\RR \\
g(x)=\left| \left| x\right| \right|-1\\
Dg=\RR\setminus\{-1;1\} \\
Zg=\RR\setminus\{-1;1\}\\
f(g(x))=f(\left| \left| x\right| \right|-1)=2(\left| \left| x\right| \right|-1)+1 \\
g(f(x))=g(2x+1)=\left| \left| 2x+1\right| \right|-1}\)
c)
\(\displaystyle{ f(x)= \sqrt{x} \\
Df=\NN \\
g(x)= x^{2} \\
Dg=\RR}\)
brak złożenia?
d)
\(\displaystyle{ f(x)=\cos x \\
Df=\RR \\
g(x)=x+1 \\
Dg=\RR\setminus\{-1\} \\
f\circ g=f(g(x))=f(x+1)=\cos (x+1) \\
g\circ f=g(f(x))=g(\cos x)=(\cos x)+1}\)
e)
\(\displaystyle{ f(x)=(x-1) ^{5} \\
Df=\RR \\
g(x)= x^{2}+3x-2 \\
f\circ g=f(g(x))=f(x^{2}+3x-2)=((x^{2}+3x-2)-1) ^{5} \\
g\circ f=g(f(x))=g((x-1) ^{5})=((x-1) ^{5})^{2}+3((x-1) ^{5})-2}\)
f)
\(\displaystyle{ f(x)= x^{2}-1 \\
g(x)= \sqrt{x-1} \\
f\circ g=f(g(x))=f(\sqrt{x-1})=(\sqrt{x-1})^{2}-1 \\
g\circ f=g(f(x))=g(x^{2}-1)=\sqrt{(x^{2}-1)-1}}\)