Question:

Step 1: Check if $f$ is one-one (injective) Assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in \mathbb{R} \setminus \{-\frac{15}{2}\}$. $$ \frac{x_1+10}{2x_1+15} = \frac{x_2+10}{2x_2+15} $$ Cross-multiplying yields: $$ (x_1+10)(2x_2+15) = (x_2+10)(2x_1+15) $$ Expanding both sides: $$ 2x_1x_2 + 15x_1 + 20x_2 + 150 = 2x_1x_2 + 15x_2 + 20x_1 + 150 $$ Subtracting $2x_1x_2 + 150$ from both sides: $$ 15x_1 + 20x_2 = 15x_2 + 20x_1 $$ Rearranging terms: $$ 20x_2 - 15x_2 = 20x_1 - 15x_1 $$ $$ 5x_2 = 5x_1 $$ $$ x_2 = x_1 $$ Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is one-one. Step 2: Check if $f$ is onto (surjective) To determine if $f$ is onto, we need to check if for every $y$ in the codomain $\mathbb{R} \setminus \{-\frac{1}{2}\}$, there exists an $x$ in the domain $\mathbb{R} \setminus \{-\frac{15}{2}\}$ such that $f(x) = y$. Let $y = f(x)$: $$ y = \frac{x+10}{2x+15} $$ Solve for $x$ in terms of $y$: $$ y(2x+15) = x+10 $$ $$ 2xy + 15y = x+10 $$ $$ 2xy - x = 10 - 15y $$ Factor out $x$: $$ x(2y-1) = 10 - 15y $$ $$ x = \frac{10 - 15y}{2y-1} $$ For $x$ to be defined, the denominator $2y-1$ must not be zero. Thus, $2y-1 \neq 0$, which implies $y \neq \frac{1}{2}$. This means that the range of $f$ is $\mathbb{R} \setminus \{\frac{1}{2}\}$. The given codomain of $f$ is $\mathbb{R} \setminus \{-\frac{1}{2}\}$. We observe that $\frac{1}{2}$ is an element of the codomain $\mathbb{R} \setminus \{-\frac{1}{2}\}$. However, from our derivation of $x$, we found that $y$ cannot be $\frac{1}{2}$. This means there is no $x$ in the domain such that $f(x) = \frac{1}{2}$. Since there exists an element in the codomain (namely $\frac{1}{2}$) that is not in the range of $f$, the function $f$ is not onto. Conclusion: The function $f$ is one-one but not onto.