📝 Analysis of Functions III (Rational Functions) in Calculus (28 MCQs)

📖 Explanation: The derivative of a polynomial is obtained by applying the power rule to each term. Differentiating $x^{3}$ gives $3x^{2}$ and differentiating $-3x^{2}$ gives $-6x$; the constant term disappears. Hence the correct derivative is $3x^{2}-6x$.

📖 Explanation: The second derivative of $f$ is f''(x)=(x-2)e^{-x}. Since the exponential factor is always positive, the sign of f'' changes where the linear factor $x-2$ equals zero, i.e., at $x=2$. Therefore the inflection point occurs at $x=2$.

📖 Explanation: A local minimum occurs where the derivative changes from negative to positive. Since f' is negative on $(0,2)$ and becomes positive after $x=2$, the sign change happens at the right endpoint of the interval, placing the minimum within $(0,2)$.

📖 Explanation: When the second derivative changes sign, the curvature of the graph switches from concave down to concave up or vice‑versa. If the first derivative is also zero, the point is not an extremum but a point where concavity changes, i.e., an inflection point.

📖 Explanation: When the first derivative passes from positive to negative, the function transitions from increasing to decreasing, which characterizes a local maximum. This reasoning holds regardless of the function’s specific form, provided the derivative exists at the point.

📖 Explanation: Writing $h(x)=x+2+\dfrac{3}{x-2}$ shows h&#039;(x)=1-\dfrac{3}{(x-2)^{2}}. The term $\dfrac{3}{(x-2)^{2}}$ is always positive, making h&#039;(x)<0 for all $x\neq2$. Hence the function is decreasing on both sides of the hole at $x=2$.

📖 Explanation: A derivative that is never negative guarantees the function never decreases. The isolated zero does not create a decrease, so the function is monotone increasing; however, because the derivative is zero at one point, the increase is not strict over the entire domain.

📖 Explanation: The sign of the second derivative indicates whether the first derivative is increasing. Since f&#039;&#039;(x)=6(x-1) is positive when $x>1$, the slope f&#039;(x) is increasing on that interval, meaning the function’s rate of increase grows for $x>1$.

📖 Explanation: On $(0,1)$ the function inherits both properties that apply to the overlapping intervals: it is part of the decreasing region $[0,2]$ and also part of the concave‑down region $(-\infty,1)$. Hence the graph is decreasing and concave down there.

📖 Explanation: The derivative of $f$ is $3x(x-2)$, giving increase on $(-\infty,0]$ and $(2,\infty)$. For $g$, g&#039;(x)=(1-x)e^{-x} is positive only when $x<1$. Thus $f$ has two separate increasing intervals, whereas $g$ has a single interval of increase.

📖 Explanation: The first derivative is f&#039;(x)=3x(x-2). Substituting $x=2$ gives f&#039;(2)=3\cdot2\cdot0=0. Hence the limit as $x$ approaches 2 of the derivative exists and equals 0.

📖 Explanation: After simplifying, $h(x)=x+2+\dfrac{3}{x-2}$. Differentiating twice yields h&#039;&#039;(x)=\dfrac{6}{(x-2)^{3}}, whose sign flips when the denominator changes sign, i.e., at the vertical asymptote $x=2$. Thus concavity changes at $x=2$.

📖 Explanation: The inflection point of a cubic occurs where its second derivative is zero. For $f$, f&#039;&#039;(x)=6(x-1) gives $x=1$. Function $m(x)=x^{3}-3x^{2}+5$ has the same second‑derivative expression, so its inflection point also occurs at $x=1$.

📖 Explanation: Total variation measures the accumulated absolute change. The sinusoidal term in $q$ causes repeated rises and falls, producing more cumulative change than the smooth cubic $p$, whose graph only climbs and falls once. Hence $q$ exhibits greater total variation on the interval.

📖 Explanation: The second derivative is f&#039;&#039;(x)=(x-2)e^{-x}. At $x=0$ this equals $(-2)e^{0}=-2$, which is negative, indicating concave down—not concave up—at the origin. Therefore the correct answer is No.

📖 Explanation: The cubic’s second derivative $6(x-1)$ vanishes once, giving a single inflection point at $x=1$. For $g$, g&#039;&#039;(x)=-2\sin x is zero at $x=0,\pi,2\pi$; interior to the interval is $\pi$. Counting interior points yields two inflection points for $g$.

📖 Explanation: Using the quotient rule, r&#039;(x)=\dfrac{f&#039;(x)g(x)-f(x)g&#039;(x)}{g(x)^{2}}. At $x=1$: f&#039;(1)=-3, $g(1)=e^{-1}$, $f(1)=-1$, g&#039;(1)=0. Numerator $(-3)e^{-1}$ is negative, denominator positive, so r&#039;(1)<0.

📖 Explanation: The second derivative quantifies how the first derivative changes. When f&#039;&#039;(x)>0, the slope is increasing, producing a “cup‑shaped” graph that is concave up. Conversely, f&#039;&#039;(x)<0 means the slope is decreasing, giving a “cap‑shaped” graph that is concave down. Hence sign of f&#039;&#039; dictates concavity.

📖 Explanation: A general cubic $ax^{3}+bx^{2}+cx+d$ has a second derivative $6ax+2b$ that vanishes at a single x‑value, guaranteeing one inflection point. Its first derivative is quadratic, which can have zero, one, or two real roots, giving up to two critical points. Thus the fourth option best captures the relationship.

📖 Explanation: From f&#039;(x)=3x(x-2) we see sign changes at $x=0$ and $x=2$, giving the stated increasing/decreasing intervals. The second derivative f&#039;&#039;(x)=6(x-1) changes sign at $x=1$, establishing concavity regions and an inflection at $(1,f(1))=(1,-1)$.

📖 Explanation: A derivative that never becomes negative ensures the function never decreases. The isolated zero does not create a decrease, so the function is monotone increasing; however, because the derivative equals zero at one point, the increase is not strict over the entire domain.

📖 Explanation: For the cubic $x^{3}-3x^{2}+1$ the inflection occurs at $x=1$. Multiplying by $e^{-x}$ as in $xe^{-x}$ introduces a damping factor that changes the second derivative to $(x-2)e^{-x}$, moving the sign‑change—and thus the inflection point—to $x=2$, i.e., to the right.

📖 Explanation: An inflection point occurs at a location where the curvature switches from concave up to concave down or vice versa. This happens precisely when the second derivative equals zero (or is undefined) and its sign changes across the point. Hence the definition relies on f&#039;&#039;.

📖 Explanation: In mathematics, a function that is concave up is also described as convex because its graph lies below its tangent lines, forming a “cup” shape. Therefore, concave up and convex are interchangeable terms, while concave down is the opposite notion.

📖 Explanation: Cancelling a common factor eliminates the discontinuity, turning a hole into a regular point. Consequently, monotonic behavior that was split by the asymptote now continues uninterrupted, causing the separate intervals of increase or decrease to merge into a single interval across the former hole.

📖 Explanation: The first derivative f&#039;(x)=(1-x)e^{-x} is positive for $x<1$ and negative for $x>1$. This sign change indicates that the function increases up to $x=1$ and then decreases, making $x=1$ the point where the function reaches its highest value on $\mathbb{R}$.

📖 Explanation: When both the first and second derivatives vanish, the test is inconclusive. If the third derivative f&#039;&#039;&#039;(x) is non‑zero, the Taylor expansion shows a change in curvature, confirming an inflection point. If f&#039;&#039;&#039;(x)=0 as well, higher‑order terms must be examined. Thus the answer depends on f&#039;&#039;&#039;(x).

📖 Explanation: An exponential growth function such as $e^{x}$ rises without bound and its second derivative is positive for all $x$, guaranteeing both increase and concave‑up behavior on the interval $(0,\infty)$.}