$ f,g\colon \mathbb{R} \to \mathbb{R}, $
$ g(f(x)) = 2x^5 + e^f + 1 $
I need to show that f(x) is 1-1
Also:
$ g(x) = x^2 -xf(x) + 1 $
show that f is not 1-1
$ f,g\colon \mathbb{R} \to \mathbb{R}, $
$ g(f(x)) = 2x^5 + e^f + 1 $
I need to show that f(x) is 1-1
Also:
$ g(x) = x^2 -xf(x) + 1 $
show that f is not 1-1
)assume that $f(x_1)=f(x_2)$, then $g(f(x_1))=g(f(x_2))$, i.e., $2x_{1}^{5}+e^{f(x_1)}+1=2x_{2}^{5}+e^{f(x_2)}+1$, then we have $2x_{1}^{5}=2x_{1}^{5}$ so $x_1=x_2$
in the second question, f(x) can be 1-1 for example f(x)=x-1