📝 Vertical tangents and cusps in graphs (25 MCQs)

📖 Explanation: The derivative follows from the power rule: differentiate $(x-4)^{2/3}$ to obtain $\frac{2}{3}(x-4)^{(2/3)-1}=\frac{2}{3}(x-4)^{-1/3}$. This matches option A, while the other choices either use the wrong exponent or coefficient.

📖 Explanation: When the derivative approaches opposite infinities from the two sides, the graph makes a sharp point where the slopes diverge. This configuration defines a cusp. A vertical tangent would require both one‑sided limits to be $+\infty$ (or both $-\infty$).

📖 Explanation: For $f(x)=(x-4)^{2/3}$, f'(x)=\frac{2}{3}(x-4)^{-1/3} which changes sign, giving $+\infty$ from the right and $-\infty$ from the left. For $g(x)=(x-4)^{1/3}$, g'(x)=\frac{1}{3}(x-4)^{-2/3} is always positive, so both limits are $+\infty$.

📖 Explanation: A vertical tangent occurs when the slope becomes infinite while the function remains continuous. Differentiability fails because the derivative does not exist as a finite number. The presence of a cusp requires opposite infinite slopes, not the same infinite slope on both sides.

📖 Explanation: The derivative h'(x)=\frac{2}{3}\operatorname{sgn}(x)\,|x|^{-1/3} approaches $+\infty$ from the right and $-\infty$ from the left, indicating opposite infinite slopes. This configuration defines a cusp at the origin.

📖 Explanation: The second derivative f''(x)=-\frac{2}{9}(x-4)^{-4/3} is negative for all $x\neq4$ because the power $-4/3$ yields a positive quantity, and the leading coefficient is negative. Hence the graph is concave down on both sides of the cusp.

📖 Explanation: A horizontal asymptote exists only when the function approaches a finite constant as $x$ grows without bound. For $y=(x-4)^{2/3}$, the expression grows like $|x|^{2/3}$, so both limits at $+\infty$ and $-\infty$ diverge to $+\infty$.

📖 Explanation: When the slope becomes infinitely large but with the same sign on both sides, the tangent line becomes vertical, producing a vertical tangent. A cusp requires opposite infinite slopes, while an inflection point involves a change in concavity without infinite slopes.

📖 Explanation: Simplifying gives $p(x)=|x|^{2/3}$. Its derivative behaves like $\frac{2}{3}\operatorname{sgn}(x)\,|x|^{-1/3}$, which tends to $+\infty$ from the right and $-\infty$ from the left, characteristic of a cusp.

📖 Explanation: A vertical tangent depends on the local behavior of the function near a specific x‑value. Translating the graph horizontally (a horizontal shift) moves the point of tangency but does not remove the vertical tangent. All other listed transformations alter the shape in ways that can eliminate the vertical tangent.

📖 Explanation: The derivative f&#039;(x)=\frac{2}{3}(x-4)^{-1/3} inherits the sign of $(x-4)^{-1/3}$. For $x<4$, the base $(x-4)$ is negative, and an odd root preserves the sign, yielding a negative derivative. For $x>4$, the derivative is positive.

📖 Explanation: q&#039;(x)=\frac{2}{3}(x-4)^{-1/3}+1. As $x\to4^{-}$, the first term $\to -\infty$ so q&#039; is negative; as $x\to4^{+}$, it $\to +\infty$ making q&#039; positive. Hence the derivative changes sign from negative to positive at $x=4$, indicating a relative minimum despite the derivative being undefined there.

📖 Explanation: An inflection point is defined by a change in the sign of the second derivative, meaning the graph switches from concave up to concave down or vice‑versa. When a vertical tangent coincides with an inflection point, the concavity on the two sides of the tangent must be opposite.

📖 Explanation: An infinite positive limit of the derivative indicates that the slope of the secant lines grows without bound, forcing the tangent line to become vertical. Continuity guarantees the point $(a,f(a))$ exists, so a vertical line through that point serves as the tangent.

📖 Explanation: Because the exponent $2/3$ is even in the numerator, raising a negative number to the power $2$ eliminates the sign before taking the cube root. Consequently $(x-4)^{2/3}=|x-4|^{2/3}$ for all real $x$; the two graphs coincide.

📖 Explanation: The derivative sign analysis shows the function decreases for $x<4$ and increases for $x>4$. This sign change indicates a transition from decreasing to increasing, which defines a relative minimum. The lack of differentiability does not prevent the minimum, as the first‑derivative test relies on sign rather than finite values.

📖 Explanation: The secant slope approaches the right‑hand derivative, which is $\displaystyle\lim_{x\to4^{+}}\frac{2}{3}(x-4)^{-1/3}=+\infty$. Hence the limit of the quotient is $+\infty$.

📖 Explanation: The function can be rewritten as $((x-4)^{1/3})^{2}$, which is an even function of the shifted variable $u=x-4$. Therefore it is symmetric about the line $u=0$, i.e., $x=4$. The answer choice stating symmetry about $x=4$ would be correct, but the options list it as D, so D is the correct answer.

📖 Explanation: A cusp implies that the first derivative approaches opposite infinities from the two sides, so the slope is unbounded. Since the second derivative involves the rate of change of this already unbounded slope, it cannot be defined at the cusp; thus it does not exist.

📖 Explanation: A vertical tangent cannot exist at a point where the function is undefined because there is no point to attach a tangent line. The other statements correctly describe properties of vertical tangents.

📖 Explanation: The limit equals the right‑hand derivative, which is $\frac{2}{3}(x-4)^{-1/3}$ as $x\to4^{+}$. This expression diverges to $+\infty$, so the limit is $+\infty$.

📖 Explanation: Adding a constant raises or lowers the entire graph without altering its shape or the x‑coordinates of any features. Consequently, the vertical tangent remains at the same $x$-value, $x=4$, regardless of the value of $k$.

📖 Explanation: Both pieces equal $|x-4|^{2/3}$, so the function is continuous at $x=4$ ($C^{0}$). However, the derivative from the left is $-\infty$ and from the right is $+\infty$, so the first derivative does not exist, meaning the function is not $C^{1}$.

📖 Explanation: The second derivative is f&#039;&#039;(x)=-\frac{2}{9}(x-4)^{-4/3}. As $x$ approaches 4 from either side, $(x-4)^{-4/3}$ tends to $+\infty$; the negative coefficient then forces the limit to $-\infty$ on both sides.

📖 Explanation: Even though the derivative does not exist at a cusp, the sign of the derivative on either side can still be examined. If the derivative is negative before the cusp and positive after, the function decreases then increases, indicating a relative minimum at the cusp.