📝 Graphing with calculus and graphing calculators (23 MCQs)

📖 Explanation: Using the quotient rule, f'(x)=\frac{x\cdot\frac{1}{x}-\ln x\cdot1}{x^{2}}=\frac{1-\ln x}{x^{2}}. This matches option B, and the other choices arise from sign or term errors that are easily spotted by careful differentiation.

📖 Explanation: Applying L'Hôpital’s rule gives $\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}= \lim_{x\to\infty}\frac{1/x}{1}=0$. Thus the horizontal asymptote is the line $y=0$. The correct answer is option D; the other alternatives confuse the limit with the numerator or suggest unbounded growth.

📖 Explanation: A relative maximum occurs where the derivative changes from positive (function increasing) to negative (function decreasing). Therefore the sign pattern is positive before the point and negative after, which corresponds to option C.

📖 Explanation: As $x$ approaches zero from the right, $\ln x$ tends to $-\infty$ while $x$ tends to $0^{+}$, making the quotient $-\infty$. Hence the limit is $-\infty$, indicating a vertical asymptote; option A reflects this correctly.

📖 Explanation: Since $\ln x<0$ when $0<x<1$ and $\ln x>0$ when $x>1$, dividing by the positive $x$ preserves the sign. Thus $f(x)$ is negative on $(0,1)$ and positive on $(1,\infty)$, matching option C.

📖 Explanation: Evaluating the second derivative at $x=e$ yields f&#039;&#039;(e)=\frac{2-3e}{e^{3}}<0, confirming concave down and thus a relative maximum. This is the second‑derivative test, making option D the correct choice.

📖 Explanation: A sign change in the second derivative indicates a transition from concave down to concave up or vice‑versa, which defines an inflection point. Therefore the graph has a point of inflection at $x=e^{3/2}$, corresponding to option C.

📖 Explanation: Setting $\frac{1-\ln x}{x^{2}}=0$ gives $\ln x=1$ so $x=e$. Evaluating f&#039;&#039;(e)=\frac{2-3e}{e^{3}}<0 shows concave down, confirming a relative maximum. Hence option B is correct.

📖 Explanation: Rewrite the limit as $\displaystyle\lim_{x\to0^{+}}\frac{1/x}{1}$ after differentiating numerator and denominator. The numerator $1/x$ diverges to $+\infty$ while the denominator stays $1$, giving $-\infty$ because $\ln x$ is negative; thus option B is correct.

📖 Explanation: Applying L’Hôpital’s rule to $\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}$ yields $\displaystyle\lim_{x\to\infty}\frac{1/x}{1}=0$. This limit confirms that the function approaches the line $y=0$ as $x$ grows, establishing the horizontal asymptote; option A captures this reasoning.

📖 Explanation: The sign of f&#039;(x)=\frac{1-\ln x}{x^{2}} is positive when $1-\ln x>0$, i.e., $\ln x<1$ or $x<e$. Since the denominator is always positive, the function increases on $(0,e)$; option A reflects this interval.

📖 Explanation: The second derivative f&#039;&#039;(x)=\frac{2\ln x-3}{x^{3}} is negative when $2\ln x-3<0$, i.e., $\ln x<\frac{3}{2}$ or $x<e^{3/2}$. Hence the graph is concave down on $(0,e^{3/2})$; option C is correct.

📖 Explanation: Setting $f(x)=0$ gives $\frac{\ln x}{x}=0$ which implies $\ln x=0$. Solving yields $x=e^{0}=1$. Therefore the intercept corresponds to the equation $\ln x=0$; option B is the correct representation.

📖 Explanation: Critical points occur where h&#039;(x)=0: at $x=1,-2,4$. The factor $(x+2)^{2}$ and $(x-4)^{2}$ indicate even multiplicity, giving no sign change, while $x=1$ (odd multiplicity) changes sign, producing a single relative extremum. Thus there is one relative extremum, option D (the fourth choice) is correct.

📖 Explanation: For $x<1$ but close, the factor $(x-1)$ is negative while the squared factors are positive, making h&#039;(x)>0. A positive derivative indicates the function is increasing just left of $x=1$; option A captures this.

📖 Explanation: Setting f&#039;&#039;(x)=0 gives $2\ln x-3=0$ → $\ln x=3/2$ → $x=e^{3/2}$. Therefore the inflection point occurs at $x=e^{3/2}$; option C is the correct answer.

📖 Explanation: Applying L’Hôpital’s rule yields $\displaystyle\lim_{x\to\infty}\frac{1/x}{1}=0$. The limit being zero shows the function approaches the line $y=0$ as $x$ grows, establishing a horizontal asymptote at $y=0$; option D states this relationship correctly.

📖 Explanation: Many utilities start plotting from a small positive value (e.g., $0.1$) to avoid division by zero, so the dramatic blow‑up near $x=0$ is not displayed. This truncation can hide the vertical asymptote, making option B the accurate description.

📖 Explanation: The stationary point occurs at $x=e$. Substituting gives $f(e)=\frac{\ln e}{e}=\frac{1}{e}$. Hence the maximum point is $(e,\frac{1}{e})$; option A provides the correct coordinate pair.

📖 Explanation: Since f&#039;&#039;(x)=\frac{2\ln x-3}{x^{3}}, the sign changes when $2\ln x-3=0$ → $x=e^{3/2}$. This value lies within the plotted interval, guaranteeing an inflection point at $x=e^{3/2}$; option C correctly identifies it.

📖 Explanation: As $x\to0^{+}$, $\ln x\to-\infty$ while $x\to0^{+}$, so the quotient tends to $-\infty$, giving a vertical asymptote. As $x\to\infty$, the quotient tends to $0$, giving a horizontal asymptote. This combination creates a graph that climbs from negative infinity near the y‑axis and approaches the x‑axis far to the right; option D captures this description.

📖 Explanation: The factor $(x^{2}-2x+4)$ is always positive, while $(x+2)^{3}$ and $(x-4)^{3}$ change sign at $-2$ and $4$. Thus h&#039;&#039;(x) changes sign at those points, indicating inflection points there, and the function is concave up on intervals where the product is positive, i.e., outside $(-2,4)$; option C reflects this.

📖 Explanation: Graphing programs sample points at discrete intervals; if the step size is larger than the distance between $2.7$ and $e$, the plotted peak appears at the nearest sampled point, giving an approximate location. Analytic calculus determines the exact stationary point by solving f&#039;(x)=0, yielding $x=e$; option A explains this discrepancy.