📝 Domain and range word problems (13 MCQs)

📖 Explanation: The side length x represents a physical cut, so it cannot be negative. The largest square that can be cut from the 16‑in side is 8 in (because 2x ≤ 16). Therefore the permissible values are all non‑negative numbers up to 8 inclusive, giving the interval $0 \le x \le 8$.

📖 Explanation: The cubic volume function first grows, attains a peak, then falls as the box becomes too shallow. Moving from 3 in (near the peak) to 7 in passes the maximum, so the volume first stops increasing and then declines toward zero, never becoming negative within the domain.

📖 Explanation: The width after cutting is $16-2x$. To keep this dimension positive, we need $16-2x > 0$, which simplifies to $x < 8$. Since x can equal 8 (giving zero width and zero volume), the domain is limited to $0 \le x \le 8$; any larger x would produce a negative width, which is physically impossible.

📖 Explanation: The variable t measures elapsed seconds starting at the exact moment 8:05 a.m., so t = 0 at the start and t = 60 at the end of the one‑minute interval. No values outside this interval are relevant to the scenario, giving the domain $0 \le t \le 60$.

📖 Explanation: $V(x)$ is a cubic polynomial that rises, reaches a peak, then falls, producing a hump‑shaped curve. In contrast, $D(t)=100t$ is a first‑degree (linear) function that forms a straight line with constant positive slope, never turning downward. Hence their shapes differ fundamentally.

📖 Explanation: The limiting dimension is the shorter side, 20 in. After cutting squares of side x from each end, the remaining width is $20-2x$. To keep this non‑negative we need $20-2x \ge 0$, giving $x \le 10$. Together with the non‑negative requirement, the domain becomes $0 \le x \le 10$.

📖 Explanation: Physical restrictions stem from real‑world constraints such as size, material strength, or safety, limiting the values a variable can take. Mathematical restrictions are imposed by the algebraic form of the function, like avoiding division by zero or taking square roots of negative numbers. The two types of limits may coincide but arise from different reasons.

📖 Explanation: The volume is the product of three positive factors: the two base dimensions and the height x. Maximum product for a fixed sum tends to occur when the factors are as close in size as possible. When x is around 3.5 in, the three dimensions become roughly balanced, yielding the greatest overall product and thus the greatest volume.

📖 Explanation: A linearly increasing speed can be expressed as $v(t)=100 + \frac{20}{60}t = 100 + \frac{1}{3}t$. Integrating speed over time gives distance: $D(t)=\int_0^t v(s)\,ds = 100t + \frac{1}{6}t^2$. Option D matches this result, reflecting the added quadratic term due to acceleration.

📖 Explanation: The range of a function consists of all output values it attains. By inspecting the graph, one can directly observe the maximum y‑coordinate (the peak of the curve) and the minimum y‑coordinate (typically zero at the domain endpoints). These observed extrema give an approximate range without performing any differentiation or algebraic solving.

📖 Explanation: Substituting $x=0$ into the volume formula $V(x) = (16-2x)(30-2x)x$ yields $V(0) = (16)(30)(0) = 0$. Physically, if no squares are removed, the box has no height, so its volume is zero. This aligns with the expectation that a dimensionless height produces no three‑dimensional space.

📖 Explanation: The volume expression $V(x) = (16-2x)(30-2x)x$ contains the factor x and also the factors $16-2x$ and $30-2x$. Setting V to zero gives x = 0 or $16-2x = 0$ (which gives x = 8) or $30-2x = 0$ (x = 15, outside the domain). Hence within the allowed domain, the volume vanishes at x = 0 and x = 8.

📖 Explanation: After removing squares of side x from each corner, the remaining width is $a-2x$. To keep this width non‑negative, we need $a-2x \ge 0$, which yields $x \le a/2$. Since a is the smaller original dimension, this condition automatically satisfies the longer side as well. Thus the domain is $0 \le x \le a/2$.