Mathematics - Functions Question with Solution | TestHub

First, let's determine the domains and ranges of $f(x)$ and $g(x)$. For $f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$ (hyperbolic tangent, $\tanh x$): Domain of $f(x)$ is $\mathbb{R}$ since $e^x+e^{-x}>0$ for all $x\in \mathbb{R}$. Range of $f(x)$: As $x\to \infty$, $f(x)\to \frac{e^x}{e^x}=1$. As $x\to -\infty$, $f(x)\to \frac{-e^{-x}}{e^{-x}}=-1$. The range of $f(x)$ is $(-1,1)$. For $g(x)=\ln \left(\frac{1+x}{1-x}\right)$:

Domain of $g(x)$: We need $\frac{1+x}{1-x}>0$. This occurs when $(1+x)$ and $(1-x)$ have the same sign. If $1-x>0⟹x<1$, then $1+x>0⟹x>-1$. So, $x\in (-1,1)$. If $1-x<0⟹x>1$, then $1+x<0⟹x<-1$, which is impossible. So, $D_g=(-1,1)$.

Range of $g(x)$: As $x\to -1^{+}$, $\frac{1+x}{1-x}\to 0^{+}$, so $g(x)\to -\infty$. As $x\to 1^{-}$, $\frac{1+x}{1-x}\to \infty$, so $g(x)\to \infty$. The range of $g(x)$ is $\mathbb{R}$.

A) The domain of $h(x)=f(g(x))$ is the set of $x$ in $D_g$ such that $g(x)$ is in $D_f$. Since $D_g=(-1,1)$ and the range of $g(x)$ is $\mathbb{R}$, which is the domain of $f(x)$, the domain of $f(g(x))$ is $D_g=(-1,1)$. This statement is correct.

B) The domain of $k(x)=g(f(x))$ is the set of $x$ in $D_f$ such that $f(x)$ is in $D_g$. Since $D_f=\mathbb{R}$ and the range of $f(x)$ is $(-1,1)$, which is the domain of $g(x)$, the domain of $g(f(x))$ is $D_f=\mathbb{R}$. This statement is correct. C) Check if $f(x)$ and $g(x)$ are odd functions:$f(-x)=\frac{e^{-x}-e^{-(-x)}}{e^{-x}+e^{-(-x)}}=\frac{e^{-x}-e^x}{e^{-x}+e^x}=-\frac{e^x-e^{-x}}{e^x+e^{-x}}=-f(x)$. So $f(x)$ is an odd function.$g(-x)=\ln \left(\frac{1+(-x)}{1-(-x)}\right)=\ln \left(\frac{1-x}{1+x}\right)=\ln \left({\left(\frac{1+x}{1-x}\right)}^{-1}\right)=-\ln \left(\frac{1+x}{1-x}\right)=-g(x)$. So $g(x)$ is an odd function. This statement is correct.

D) The range of $f(x)$ is $(-1,1)$ and the range of $g(x)$ is $\mathbb{R}$. This statement incorrectly swaps the ranges. This statement is incorrect.