📝 Absolute Maxima and Minima in calculus (24 MCQs)

📖 Explanation: By the Extreme Value Theorem, a continuous function on a closed interval attains both an absolute maximum and an absolute minimum somewhere in the interval. The theorem does not specify that the extrema occur at the endpoints, only that they exist within the interval.

📖 Explanation: Because $g$ is decreasing throughout $(a,b)$, the function’s values get smaller as $x$ increases. A stationary point where the derivative is zero cannot be a local extreme for a strictly decreasing function, so it cannot be an absolute extremum; the extrema must occur at the interval endpoints.

📖 Explanation: Differentiating gives h'(x)=\frac{1-x^{2}}{(1+x^{2})^{2}}, which vanishes at $x=\pm1$. Evaluating $h$ at these points yields $h(\pm1)=\pm\frac12$. However, the true extrema occur where h'(x)=0 and the second derivative test shows maxima at $x=\frac{1}{\sqrt{3}}$ and minima at $-\frac{1}{\sqrt{3}}$ with values $\pm\frac{2}{3\sqrt{3}}$. Thus option B is correct.

📖 Explanation: On $[-2,3]$, $f$ reaches its maximum at $x=3$ with value $9$. The function $g$ reaches its maximum at the same endpoint $x=3$ with value $27$. Since $27>9$, $g$ has the larger absolute maximum on the given interval.

📖 Explanation: Compute p'(x)=3x^{2}-3=3(x^{2}-1). Critical points are at $x=\pm1$. Evaluating $p$ at endpoints and critical points gives $p(-2)=-2$, $p(-1)=2$, $p(1)=-2$, $p(2)=2$. The largest value is $2$ occurring at $x=2$ and also at $x=-1$; the smallest value is $-2$ at $x=-2$ and $x=1$. Hence the absolute maximum is at $x=2$ and the absolute minimum at $x=-2$.

📖 Explanation: On $[0,1]$, $k(x)=x^{2}$ attains its largest value $1$ at $x=1$. For $x>1$, $k(x)=2x+1$ is increasing, reaching $k(3)=7$ at the right endpoint. The smallest value on the entire interval is $k(0)=0$. Hence the absolute maximum occurs at the endpoint $x=3$ and the absolute minimum at $x=0$.

📖 Explanation: The theorem states that if a function is continuous on a closed, bounded interval $[a,b]$, then it must achieve both a greatest and a least value somewhere in that interval. Openness or unboundedness can allow the function to approach but never reach its supremum or infimum, violating the conclusion of the theorem.

📖 Explanation: The natural logarithm is strictly increasing. Therefore the smallest value on $[1,e^{2}]$ occurs at the left endpoint $x=1$ where $\ln(1)=0$, and the largest value occurs at the right endpoint $x=e^{2}$ where $\ln(e^{2})=2$. Hence the function has an absolute minimum at $x=1$ and an absolute maximum at $x=e^{2}$.

📖 Explanation: The function $e^{-x^{2}}$ is always positive and attains its largest value when the exponent $-x^{2}$ is zero, i.e., at $x=0$, giving a value of 1. As $|x|$ grows, the exponent becomes large negative, causing the function to approach 0 but never reach it, so there is no absolute minimum.

📖 Explanation: Compute f'(x)=4x^{3}-8x=4x(x^{2}-2). Critical points are at $x=0,\pm\sqrt{2}$. Evaluating $f$ at endpoints and critical points yields $f(0)=0$, $f(\pm\sqrt{2})=-4$, and $f(\pm3)=-27$. The largest value is $0$ at $x=0$, making it the absolute maximum.

📖 Explanation: On $[0,\pi]$, $\sin x$ increases from 0 to 1 at $x=\frac{\pi}{2}$ and then decreases back to 0 at $x=\pi$. Hence the absolute maximum value $1$ occurs at $x=\frac{\pi}{2}$, while the absolute minimum value $0$ occurs at both endpoints, $x=0$ and $x=\pi$. Option D captures the maximum and one of the minima.

📖 Explanation: A positive second derivative at a critical point indicates a local minimum. Since $p$ has only one critical point in the interval and the function is continuous, the absolute minimum must occur at that point if the endpoint values are larger. Thus $x=1$ is the absolute minimum on $[0,2]$.

📖 Explanation: For $x\le0$, $q(x)=x+2$ increases to its largest value $2$ at $x=0$. For $x>0$, $-x^{2}+2$ decreases from $2$ at $x=0$ to $-2$ at $x=2$. Hence the highest value of the function on $[-2,2]$ is $2$ at $x=0$; the lowest value is $-2$ at $x=2$. Option A correctly identifies the absolute maximum.

📖 Explanation: On $[1,5]$, $\frac{1}{x}$ is decreasing. Therefore the largest value occurs at the left endpoint $x=1$ where $r(1)=1$, and the smallest value occurs at the right endpoint $x=5$ where $r(5)=0.2$. Hence the function possesses an absolute maximum at $x=1$.

📖 Explanation: First find critical points: s'(x)=3x^{2}-12x+9=3(x^{2}-4x+3)=3(x-1)(x-3). Critical points in $[0,4]$ are $x=1$ and $x=3$. Evaluating $s$ at endpoints and critical points yields $s(0)=0$, $s(1)=4$, $s(3)=0$, $s(4)=4$. The smallest value is $0$ occurring at both $x=0$ and $x=3$; thus the absolute minimum is attained at $x=0$ (or $x=3$). Option C is incorrect; the correct answer is $x=0$. However, based on the options, $x=0$ (Option A) is the absolute minimum.

📖 Explanation: A negative second derivative at a critical point indicates a local maximum. However, without comparing the function values at the endpoints, we cannot guarantee that this local maximum is the absolute maximum on the closed interval. Hence the only certain conclusion is that $c$ is a local maximum; it may or may not be absolute.

📖 Explanation: The function represents the upper half of a circle of radius 3 centered at the origin. Its largest value is the radius itself, achieved at $x=0$ where $u(0)=3$. At the endpoints $x=\pm3$, the function value is zero, giving the absolute minimum. Hence option A is correct.

📖 Explanation: An absolute maximum at $x_{0}$ means that the function’s value at that point is greater than or equal to its value at every other point in the entire interval. This differs from a relative (local) maximum, which only requires the inequality to hold in a neighborhood around $x_{0}$.

📖 Explanation: The cosine function reaches its highest value $1$ at $x=0$. Its lowest value $-1$ occurs at both endpoints $x=\pi$ and $x=-\pi$. Therefore the absolute maximum is at $x=0$ and the absolute minimum occurs at both $x=\pm\pi$. Option D captures this correctly.

📖 Explanation: Complete the square: $w(x)=(x+2)^{2}+3$. The vertex at $x=-2$ yields the minimum value $3$. Since $-2$ lies within $[-5,1]$, the absolute minimum occurs at $x=-2$.

📖 Explanation: Differentiability on an open interval does not preclude the existence of absolute extrema; for example, $f(x)=x^{2}$ on $(-1,1)$ has an absolute minimum at $x=0$ even though the interval is open. Thus statement C is false.

📖 Explanation: Since $x^{2}+1$ is increasing for $x\ge0$, the logarithm of this quantity also increases. Therefore the largest value occurs at the right endpoint $x=2$, where $y(2)=\ln(5)$.

📖 Explanation: The expression simplifies to a value between 0 and 1 for all real $x$. As $|x|$ grows, the fraction approaches 1 but never exceeds it, so the supremum is 1, not attained. At $x=0$, the value is 0, giving the absolute minimum. Hence option C is correct.

📖 Explanation: A positive derivative on the open interval indicates that the function is strictly increasing. Consequently, the smallest value occurs at the left endpoint $a$ and the largest at the right endpoint $b$. Thus the absolute minimum is at $a$ and the absolute maximum at $b$.