📝 Relative maxima and minima local extrema (24 MCQs)

📖 Explanation: A relative maximum is a point where the function’s value is at least as large as every other value in a small neighbourhood around that point. This matches option A, which states the inequality condition for all \x\ in an open interval containing \x_{0}\. The other options describe minima or derivative conditions, not the definition of a relative maximum.

📖 Explanation: A critical point occurs wherever the first derivative is zero (a stationary point) or fails to exist (a cusp, corner, or vertical tangent). Option C captures both possibilities, while the other options describe unrelated situations such as discontinuities or absolute extrema.

📖 Explanation: The derivative is \f'(x)=3x^{2}-3=3(x^{2}-1)\, giving critical points at \x=\\pm1\. Sign analysis shows the function increases before \-1\, decreases between \-1\ and \1\, then increases after \1\. Thus a relative maximum occurs at \x=-1\ and a relative minimum at \x=1\, which is option B.

📖 Explanation: When the first derivative switches from positive (function increasing) to negative (function decreasing), the function reaches a high point in its local neighbourhood, which is precisely the definition of a relative maximum. Therefore option A correctly describes the outcome of this sign change.

📖 Explanation: When the derivative is zero yet retains the same sign on both sides of \c\, the function continues either increasing or decreasing through \c\. Hence there is no local high or low point, and \c\ is not a relative extremum. Option D accurately reflects this situation.

📖 Explanation: The derivative \f'(x)=4x^{3}-8x=4x(x^{2}-2)\ gives a critical point at \x=0\. The second derivative \f''(x)=12x^{2}-8\ evaluates to \-8\ at \x=0\, indicating concave down, but the sign of \f'\ does not change (it is zero on both sides). Therefore \x=0\ is not a relative extremum, matching option C.

📖 Explanation: Relative maxima of \\\sin x\ occur where the derivative \\\cos x\ changes from positive to negative, which happens at \x=\\frac{\\pi}{2}\ and again at \x=\\frac{5\\pi}{2}\ outside the given interval. Within \[0,2\\pi]\ the only such point is \x=\\frac{\\pi}{2}\. Option B incorrectly includes \\\frac{3\\pi}{2}\ (a minimum), so the correct answer is actually A. However, based on the provided options, the best match is A, which lists the true maxima within the broader domain.

📖 Explanation: A stationary point is defined by \f'(x)=0\. While many relative extrema occur at stationary points, some stationary points are inflection points (saddle points) where the function changes concavity without a local max or min. Hence a stationary point can be either a relative extremum or not, making option C correct.

📖 Explanation: The polynomial can be rewritten as \(x-1)^{4}\, which has a single critical point at \x=1\. The derivative is zero there, but the sign of \h'\ does not change, so there is no relative maximum or minimum. Therefore the function has zero relative extrema, making option A the correct answer. However, based on the options provided, the closest answer is C (two), which is inaccurate. The accurate answer is zero.

📖 Explanation: First compute \p'(x)=3x^{2}-12x+9=3(x^{2}-4x+3)=3(x-1)(x-3)\. Critical points are \x=1\ and \x=3\. The second derivative \p''(x)=6x-12\ gives \p''(3)=6\, which is positive, indicating concave up, so \x=3\ is a relative minimum. Thus option B is correct, not C. The correct answer is B.

📖 Explanation: A cusp means the derivative does not exist, so \x=2\ is a critical point. Whether it is a relative extremum depends on the surrounding behavior of the function. Therefore, option D correctly states that the point is critical and may be an extremum, while the other options make unwarranted assertions.

📖 Explanation: The function \f(x)=x^{2}\ is a parabola opening upward, and its derivative \f'(x)=2x\ is zero at \x=0\. The second derivative \f''(x)=2>0\ confirms a concave‑up shape, indicating a relative (and absolute) minimum at the origin. None of the other choices produce a minimum at \x=0\.

📖 Explanation: At a cusp the derivative does not exist, but the function may still change from decreasing to increasing, creating a local low point. This behavior satisfies the definition of a relative minimum even though the standard derivative test cannot be applied. Option A captures this reasoning.

📖 Explanation: A relative maximum occurs when the slope is zero (horizontal tangent) and the concavity is downward. The first‑derivative condition \f&#039;(c)=0\ gives a stationary point, and the second‑derivative condition \f&#039;&#039;(c)<0\ ensures the graph is curving downward, confirming a local maximum. Hence option A is correct.

📖 Explanation: For \x\\le0\, \q(x)=x^{2}\ has a minimum at \x=0\. For \x>0\, \q(x)=-x\ is decreasing, so the value just to the right of zero is negative, lower than \q(0)=0\. Hence \x=0\ is not a minimum. However, the left‑hand side derivative is zero while the right‑hand side derivative is \-1\, indicating a cusp with a change from decreasing to increasing, making \x=0\ a relative maximum. Therefore option B is correct.

📖 Explanation: Even functions satisfy \f(x)=f(-x)\. Consequently, any relative extremum at \x=a\ must be mirrored at \x=-a\. This symmetry about the \y\-axis means that relative maxima and minima occur in symmetric pairs, making option C the accurate statement.

📖 Explanation: For \f(x)=\\cos x\, \f&#039;(x)=-\\sin x\ and \f&#039;&#039;(x)=-\\cos x\. At \x=2k\\pi\ (even multiples of \\\pi\), \f&#039;(x)=0\ and \f&#039;&#039;(x)=-1<0\, giving relative maxima. At \x=(2k+1)\\pi\ (odd multiples), \f&#039;(x)=0\ and \f&#039;&#039;(x)=1>0\, giving relative minima. Thus both statements are true, so option C is correct.

📖 Explanation: When the derivative is positive before a point and negative after, the function rises up to that point and then falls, indicating a local high point—a relative maximum. Therefore, the correct inference is a relative maximum at \x=3\, corresponding to option A.

📖 Explanation: Compute \r&#039;(x)=5x^{4}-15x^{2}+4=5(x^{4}-3x^{2})+4\. Solving \r&#039;(x)=0\ yields three real critical points (approximately \-2,0,2\). Sign analysis shows the derivative changes sign at each, producing two relative extrema (a maximum and a minimum). Hence the correct answer is two, which corresponds to option B. However, given the options, the closest is C (three), which is inaccurate.

📖 Explanation: The first‑derivative test relies solely on the sign of \f&#039;\ around a critical point and does not require the existence of \f&#039;&#039;\. When the second derivative is undefined or difficult to compute (e.g., at cusps), the first‑derivative test can still determine the presence of a relative extremum. Hence option A correctly captures this advantage.

📖 Explanation: For \x\\le1\, \s(x)=x^{2}\ has its minimum at \x=0\ with value 0. At \x=1\, the left‑hand limit gives \s(1)=1\ while the right‑hand value is \2(1)-1=1\; the function is continuous. The derivative from the left is \2x\ (zero at \x=0\), and from the right is 2, indicating a change from decreasing to increasing at \x=0\. Therefore \x=0\ is the relative minimum, making option A correct.

📖 Explanation: An odd function satisfies \f(-x)=-f(x)\. Consequently, any relative extremum at \x=a\ has a counterpart at \x=-a\ with opposite function value, creating symmetry about the origin. This property makes option C the correct description of relative extrema for odd functions.

📖 Explanation: The derivative is \f&#039;(x)=\\frac{2x}{x^{2}+1}\, which is zero at \x=0\. The second derivative \f&#039;&#039;(x)=\\frac{2(1-x^{2})}{(x^{2}+1)^{2}}\ evaluates to 2>0 at \x=0\, indicating concave up and thus a relative (and absolute) minimum. Therefore option B is correct.

📖 Explanation: When the first non‑zero derivative after the first derivative is of even order and positive, the point is a relative minimum. Here the second and third derivatives vanish, while the fourth derivative is positive, satisfying the condition for a local minimum. Hence \c\ is a relative minimum, making option B correct.