Q.A function is defined by $f(x) = 2x+5$ for $x \in \mathbb{R}$. Find the value of $x$ for which $f(x)=0$, and state the domain and range of $f$.

This is a linear function, defined for every real $x$ with no restriction, so we first identify domain/range from its type, then solve the given equation.

Since $f(x)=2x+5$ is a polynomial (linear), it is defined for every real $x$, so the domain is $\mathbb{R}$; and as $x$ ranges over all reals, $2x+5$ also takes every real value, so the range is $\mathbb{R}$. Setting $f(x)=0$: $2x+5=0 \Rightarrow x=-\dfrac52$.

Step 1 — Domain and range. $f(x)=2x+5$ is a linear polynomial; polynomials involve no division and no roots, so there is no value of $x$ that must be excluded — Domain $=\mathbb{R}$. Since the coefficient of $x$ is non-zero ($2\neq0$), the line is neither horizontal nor vertical, so as $x$ sweeps through all real numbers, $y=2x+5$ also sweeps through all real numbers — Range $=\mathbb{R}$.

Step 2 — Solve $f(x)=0$. $2x+5=0 \Rightarrow 2x=-5 \Rightarrow x=-\dfrac52$.

Check (independent verification). Substitute back: $f\!\left(-\dfrac52\right) = 2\left(-\dfrac52\right)+5 = -5+5=0$ ✓, confirming the solution. For the range claim, note $f\!\left(\dfrac{y-5}{2}\right)=y$ for any target value $y$, showing every real $y$ is indeed attained.

A frequent error is a sign slip when isolating $x$ (writing $x=\dfrac52$ instead of $-\dfrac52$). Students also sometimes assume every function must have a restricted domain and needlessly search for excluded values in a plain linear function, where none exist.