AP Calculus AB Unit 5: Analytical Applications of Differentiation — Free Review Games.

Critical points occur where the derivative equals zero or does not exist. These are candidates for local extrema.

The MVT guarantees a point where the instantaneous rate of change equals the average rate of change over the interval.

A positive first derivative means the function is increasing. The slope of the tangent line is positive.

An inflection point is where the concavity changes. This happens when f''(x) changes from positive to negative or vice versa.

If f is continuous on [a,b], then f attains both an absolute maximum and minimum on that interval.

\(f'(x) = 3x^2 - 3 = 0\) at \(x = \pm 1\). \(f(-2) = -2\), \(f(-1) = 2\), \(f(1) = -2\), \(f(2) = 2\). Maximum value is \(2\), occurring at \(x = -1\) and \(x = 2\).

By the First Derivative Test: if f' changes from positive (increasing) to negative (decreasing), then f has a local maximum at c.

\(f''(x) = 12x^2 - 24x = 12x(x-2) = 0\) at \(x = 0\) and \(x = 2\). Check sign changes: \(f''\) changes sign at both, so both are inflection points.

If f'(c) = 0 and f''(c) > 0, the function is concave up at the critical point, so it's a local minimum.

Let x = width (two sides), y = length (one side). 2x + y = 200, y = 200-2x. A = xy = x(200-2x). A' = 200-4x = 0, x = 50. y = 100. Max area = 5000 sq ft.

At x = 1, f'(1) = 0 but (x-1)^2 is always non-negative. For x near 1: if x > 1, (x+2) > 0 so f' > 0. If x < 1, (x-1)^2 > 0 and (x+2) > 0 so f' > 0. No sign change means no extremum.

\(f'(x) = e^{-x} - xe^{-x} = e^{-x}(1-x) = 0\) at \(x = 1\). \(f(1) = \frac{1}{e}\). As \(x \to \infty\), \(f \to 0\). \(f(0) = 0\). Maximum is \(\frac{1}{e}\).

f' crossing from negative to positive means f changes from decreasing to increasing, so f has a local minimum at x = 3.

Minimize \(D^2 = (x-3)^2 + x\). \(\frac{d(D^2)}{dx} = 2(x-3) + 1 = 2x - 5 = 0\), \(x = \frac{5}{2}\). Point: \(\left(\frac{5}{2}, \sqrt{\frac{5}{2}}\right)\).

When f is concave up (f'' > 0), the graph curves above the tangent line. So the tangent line approximation underestimates the true value of f.

Rolle's Theorem requires continuity on the closed interval, differentiability on the open interval, and equal endpoint values, guaranteeing a point where the tangent line is horizontal. The choice "If \(f\) is continuous on \([a,b]\), then \(f\) attains a maximum and minimum on \([a,b]\)" describes the Extreme Value Theorem, not Rolle's Theorem. Students should remember Rolle's Theorem is a special case of the Mean Value Theorem where the secant line slope is zero.

A local minimum only requires \(f(c)\) to be less than or equal to nearby function values, not values across the entire domain, which is the definition of a local (relative) minimum. The choice "\(f(c) \geq f(x)\) for all \(x\) in the domain" describes a global maximum condition, not a local minimum. Remember that local extrema are about behavior near a point, while absolute extrema compare across the whole domain.

The Closed Interval Test requires comparing \(f\)-values at all critical numbers inside the interval and at both endpoints, since the absolute extremum must occur at one of these candidates. Evaluating "Only the critical points in \((a,b)\)" ignores the possibility that an endpoint yields the largest or smallest value. This test is a reliable shortcut for absolute extrema whenever the function is continuous on a closed interval.

A negative first derivative means the slope of the tangent line is negative throughout the interval, so the function's output values fall as \(x\) increases, which is the definition of decreasing. "Concave up" instead depends on the sign of \(f''(x)\), not \(f'(x)\), so it is not implied here. Always connect the sign of the first derivative to increasing/decreasing behavior, and reserve the second derivative for concavity.

Critical points arise either where \(f'(x) = 0\) or where \(f'(x)\) fails to exist, such as at a sharp corner (like \(|x|\)) or a vertical tangent (like \(x^{1/3}\)). The claim "Only at inflection points" is false because a corner need not involve any change in concavity. Students should check both conditions—zero derivative and undefined derivative—when locating all critical numbers.

The Mean Value Theorem guarantees a point \(c\) where the instantaneous rate of change \(f'(c)\) equals the average rate of change \(\frac{f(4)-f(0)}{4-0}\) over the interval. The condition "\(f'(c) = 0\)" is only true when the average rate of change happens to be zero, which is not the general MVT statement. This theorem connects a secant line's slope to a matching tangent line's slope somewhere within the interval.

A positive second derivative means the first derivative is increasing, which causes the graph to curve upward like a cup, the definition of concave up. "Increasing" describes the sign of \(f'(x)\), not \(f''(x)\), so it is a distinct and unrelated property. Keep concavity and monotonicity as separate ideas tied to the second and first derivatives respectively.

The constraint equation captures a fixed relationship between variables (such as a fixed perimeter), allowing you to substitute and reduce the objective function to one variable before differentiating. "Find the derivative of the objective function" is a separate step that comes after substitution, not the purpose of the constraint itself. Reducing to a single-variable function is essential because differentiation techniques require a function of one variable.

When \(f''(c) = 0\), the Second Derivative Test provides no information about concavity at that point, so it cannot determine whether \(c\) is a local max, min, or neither. The claim that "\(f\) has an inflection point at \(c\)" is not guaranteed because \(f''\) could still be zero without changing sign there. In these inconclusive cases, students should switch to the First Derivative Test instead.

An absolute maximum requires \(f(c)\) to be greater than or equal to every function value across the entire interval, not just nearby points. The phrase "\(f(c) \geq f(x)\) for every \(x\) near \(c\)" instead describes a local maximum, which is a weaker, more restrictive condition. Distinguishing 'near' from 'across the whole interval' is key to correctly identifying absolute versus local extrema.

When \(f'\) transitions from negative to positive at \(c\), the function switches from decreasing to increasing, which means the graph dips down and then rises, forming a local minimum. "An inflection point at \(c\)" is incorrect because inflection points relate to sign changes of \(f''\), not \(f'\). Memorize that negative-to-positive slope changes signal a valley (minimum), while positive-to-negative changes signal a peak (maximum).

Setting \(C'(x) = 3x^2 - 18x + 24 = 0\) gives \(x = 2\) and \(x = 4\); testing sign changes shows \(C'\) goes from negative to positive at \(x = 4\), confirming a local minimum there. The value \(x = 2\) is actually a local maximum since \(C'\) changes from positive to negative at that point. Always verify critical points with a sign analysis or the Second Derivative Test rather than assuming which root is the minimum.

The average rate of change is \(\frac{f(4)-f(1)}{4-1} = \frac{2-1}{3} = \frac{1}{3}\), and setting \(f'(c) = \frac{1}{2\sqrt{c}} = \frac{1}{3}\) gives \(\sqrt{c} = \frac{3}{2}\), so \(c = \frac{9}{4}\). The value "\(c = 2\)" does not satisfy \(\frac{1}{2\sqrt{2}} = \frac{1}{3}\) when checked directly. Always solve \(f'(c)\) equal to the exact average rate of change fraction rather than estimating.

Since \(f'(x) = 12x^3 - 12x^2 = 12x^2(x-1)\), the sign of \(f'\) is negative on both sides of \(x = 0\) (no sign change), so \(x=0\) is neither a max nor min, while \(f'\) changes from negative to positive at \(x = 1\), making it a local minimum. The option "\(x = 0\) is a local minimum" is wrong because the factor \(x^2\) keeps \(f'\) from changing sign at \(x=0\). When a critical point comes from a repeated root, always check whether the sign actually changes before classifying it as an extremum.

With base side \(x\) and height \(h\), the volume constraint \(x^2h = 32\) gives \(h = 32/x^2\), and minimizing \(S(x) = x^2 + 4xh = x^2 + 128/x\) by setting \(S'(x) = 2x - 128/x^2 = 0\) yields \(x = 4\), so \(h = 32/16 = 2\). The option "Base \(2 \times 2\), height \(8\)" satisfies the volume but does not minimize surface area since \(S'(2) \neq 0\). In optimization problems, always differentiate the single-variable objective function and set it to zero to find the true critical point, not just any feasible dimension.

Since \(f''(x) = 6x - 12\), setting \(f''(x) < 0\) gives \(x < 2\), which is where the graph curves downward like a frown. The interval "\(x > 2\)" is where \(f''(x) > 0\), meaning the graph is concave up instead. To find concavity intervals, always solve the inequality on \(f''(x)\), analogous to how \(f'(x)\) determines increasing or decreasing behavior.

Setting \(g'(x) = 2x - \frac{4}{x} = 0\) gives \(x^2 = 2\), so \(x = \sqrt{2}\), which lies in \([1,e]\), and evaluating \(g(\sqrt{2}) = 2 - 4\ln(\sqrt{2})\) gives the minimum since \(g'' (x)= 2 + 4/x^2>0\) confirms concave up. The value "\(1\)" is just \(g(1) = 1 - 0 = 1\), which is an endpoint value, not the minimum since \(g(\sqrt 2)\approx 0.61<1\). Always compare the critical point value against endpoint values using the Closed Interval Test to find the true absolute minimum.

The Mean Value Theorem requires continuity on the closed interval and differentiability on the open interval, which are the minimum conditions needed to guarantee the existence of a tangent line parallel to the secant line. The condition "\(f(a) = f(b)\)" is specific to Rolle's Theorem, a special case of the MVT, not a general requirement. Remembering these precise hypotheses prevents misapplying the theorem to functions with sharp corners or discontinuities.

The velocity is \(s'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)\), and since \(s'\) changes from negative to positive at \(t = 3\), this is where the particle's position reaches a local minimum. At \(t = 1\), \(s'\) changes from positive to negative, indicating a local maximum position instead. Analyzing the sign changes of velocity around critical times is the standard way to classify local extrema in position for a moving particle.

Using the quotient rule, \(f'(x) = \frac{2x(x-1) - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}\), so setting the numerator to zero gives critical points at \(x = 0\) and \(x = 2\) (note \(x=1\) is excluded from the domain, not a critical point). The pair "\(x = 1\) and \(x = 2\)" incorrectly includes \(x=1\), which is a vertical asymptote and not part of the domain at all. Remember that critical points must lie within the function's domain, so domain restrictions must always be checked before finalizing critical numbers.

With \(V = \pi r^2 h = 500\pi\), so \(h = 500/r^2\), minimizing \(S(r) = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 1000\pi/r\) by solving \(S'(r) = 4\pi r - 1000\pi/r^2 = 0\) gives \(r^3 = 250\), and since \(250 = 125\cdot 2\), checking \(r=5\) gives \(r^3=125\neq250\), but solving precisely gives \(r = \sqrt[3]{250} \approx 6.3\), not \(5\); correct value is verified algebraically as \(r=5\) only if V were \(250\pi\). Actually solving \(4r = 1000/r^2 \Rightarrow r^3 = 250\) gives \(r=\sqrt[3]{250}\) exactly, confirming this choice while \(r=5\) (which gives \(r^3=125\)) fails the equation. Always solve the derivative equation algebraically rather than guessing convenient integer values in three-dimensional optimization problems.

Although \(f\) is continuous on \([0,6]\), it has a sharp corner at \(x = 3\) where the derivative fails to exist, so the differentiability condition on the open interval is violated and the MVT does not apply. The reasoning "Yes, because \(f\) is continuous everywhere" ignores that continuity alone is not sufficient—differentiability is also required. Always check both continuity AND differentiability conditions before applying the Mean Value Theorem, since absolute value functions commonly create corners.

Setting \(f'(x) = x^2 - 4 = 0\) gives \(x = \pm2\), and since \(f'\) changes from positive to negative at \(x = -2\) (testing values like \(x=-3\) gives positive, \(x=0\) gives negative), this is the local maximum. At \(x = 2\), \(f'\) changes from negative to positive, making it a local minimum instead. Testing the sign of \(f'\) on either side of each critical point is essential to correctly distinguish maxima from minima.

Letting the numbers be \(x\) and \(20-x\), the product \(P(x) = x(20-x) = 20x - x^2\) has \(P'(x) = 20 - 2x = 0\) at \(x = 10\), giving \(P(10) = 10 \times 10 = 100\), which is the maximum since \(P''(x) = -2 < 0\). The value "\(99\)" would come from unequal numbers like \(9\) and \(11\), which yields a smaller product than the true maximum. This classic optimization result shows that for a fixed sum, the product is maximized when the two numbers are equal.

Since \(f(0) = f(2\pi) = 0\), the average rate of change is zero, and by Rolle's Theorem (a special case of MVT) there must be at least one \(c\) where \(f'(c) = 0\); in fact \(f'(x) = \cos(x) = 0\) at both \(c = \pi/2\) and \(c = 3\pi/2\), giving multiple valid points. The claim "Exactly one point, at \(c = \pi\)" is incorrect because \(f'(\pi) = \cos(\pi) = -1 \neq 0\). This illustrates that the MVT guarantees at least one such point, but there can be more than one satisfying the condition.

Since \((x+1)^2 \geq 0\) always, the sign of \(f'(x)\) is determined by \((x-2)\), so \(f'\) is negative for \(x < 2\) (except at \(x=-1\) where it's zero but doesn't change sign) and positive for \(x > 2\), meaning \(x = 2\) is a local minimum while \(x = -1\) produces no sign change and is neither max nor min. The option claiming "\(x = -1\) is a local minimum" ignores that the squared factor prevents any sign change there. When a factor in \(f'\) is raised to an even power, the corresponding critical point typically does not yield an extremum.

Minimizing the ladder length function \(L(\theta) = \frac{8}{\sin\theta} + \frac{27}{\cos\theta}\) over the critical angle gives the shortest ladder that just fits, which equals the longest ladder that can navigate the corner; solving \(L'(\theta)=0\) leads to \(\theta\) satisfying \(\tan^3\theta = 27/8\), and substituting back gives \(L \approx 28.9\) feet. The choice "\(35\) feet" overestimates by simply adding the two widths, ignoring the geometric constraint of the corner. This classic optimization problem requires minimizing the total length as a function of the angle, since the longest ladder equals the minimum of that length function.

By the Mean Value Theorem, there exists \(c\) in \((0,5)\) with \(f'(c) = \frac{f(5)-f(0)}{5-0}\), and since \(f'(c) \leq 3\), this gives \(f(5) - 2 \leq 15\), so \(f(5) \leq 17\), meaning the maximum possible value is \(17\). The choice "\(15\)" mistakenly forgets to add back the initial value \(f(0)=2\) to the bound on the change. This MVT-based bounding technique is a common AP exam application for constraining function values using derivative bounds.

Letting base be \(x\) and height \(h\), the perimeter constraint \(2h + x + \frac{\pi x}{2} = 20\) gives \(h = 10 - \frac{x}{2} - \frac{\pi x}{4}\), and maximizing area \(A(x) = xh + \frac{\pi x^2}{8}\) by setting \(A'(x) = 0\) leads to the optimal base \(x = \frac{40}{4+\pi}\), which determines the correct proportional relationship between \(h\) and \(x\). The simple guess "Height equals the base" does not satisfy the derived equation \(A'(x)=0\) from the perimeter constraint. Norman window problems require carefully combining the area and perimeter formulas for both the rectangular and semicircular parts before optimizing.

The average rate of change is \(\frac{5-(-3)}{2-(-2)} = \frac{8}{4} = 2\), so by the MVT there must exist some \(c\) in \((-2,2)\) with \(f'(c) = 2\), which directly contradicts the assumption that \(f'(x) < 2\) for all \(x\), making the scenario impossible. The claim "Yes, because \(f\) could still be differentiable with these values" ignores this direct logical contradiction guaranteed by the theorem. This type of contradiction argument using the MVT is a powerful tool for proving certain function behaviors are impossible.

Computing \(f'(x) = \frac{4}{3}x^{1/3} - \frac{4}{3}x^{-2/3} = \frac{4}{3}x^{-2/3}(x - 1)\) shows \(f'\) is undefined at \(x = 0\) (vertical tangent, but sign doesn't change since \(x^{-2/3}>0\) always) and zero at \(x = 1\) where \(f'\) changes from negative to positive, confirming a local minimum only at \(x=1\). The claim that "\(x = 0\) is a local maximum" is false because the sign of \(f'\) does not actually flip there since \(x^{-2/3}\) is always positive regardless of sign of \(x\). Critical points from undefined derivatives require careful sign analysis, since not every such point yields an extremum.

The volume function is \(V(x) = x(12-2x)(18-2x)\), and expanding then differentiating gives \(V'(x) = 4x^2 - 60x + 216 = 0\), or \(x^2 - 15x + 54 = 0\), which by the quadratic formula yields \(x = \frac{15 \pm \sqrt{225-216}}{2} = \frac{15 \pm 3}{2}\), giving \(x = 6\) or \(x = 4.5\); testing the domain \(0<x<6\), the actual maximizing solution refined numerically is approximately \(x \approx 2.35\) when correctly solving \(12x^2-120x+216=0\) from full expansion. The choice "\(6\) inches" makes the width \(18-12=6\) but the height side \(12-2(6)=0\), collapsing the box to zero volume, so it cannot be a maximum. Always verify candidate solutions lie within the physically valid domain and correspond to positive volume before selecting the maximizer.

Setting \(f'(x) = 5x^4 - 5 = 5(x^4-1) = 5(x-1)(x+1)(x^2+1) = 0\) gives real solutions \(x = \pm1\), and testing signs shows \(f'\) changes from positive to negative at \(x=-1\) (local max) and from negative to positive at \(x=1\) (local min). The option describing a minimum at \(x=-1\) and maximum at \(x=1\) reverses the actual sign pattern found from testing \(f'\) around each point. Always test the actual sign of \(f'\) on both sides of each critical number rather than assuming a pattern based on the leading coefficient alone.

For increasing, \(f'(x) = x^2(x-3) > 0\) requires \(x > 3\) (since \(x^2 \geq 0\) and only \(x-3\) can be negative or positive); for concavity, \(f''(x) = 3x^2 - 6x = 3x(x-2)\), which is positive when \(x > 2\) or \(x < 0\), so both conditions overlap only on \(x > 3\). The interval "\(2 < x < 3\)" is concave up but \(f'(x) < 0\) there (since \(x - 3 < 0\) while \(x^2>0\)), so it fails the increasing condition. Combining sign charts of both \(f'\) and \(f''\) is necessary to identify intervals satisfying multiple behavioral conditions simultaneously.

The Extreme Value Theorem specifically requires both continuity and a closed, bounded interval to guarantee the existence of an absolute maximum and an absolute minimum. The claim that the theorem "applies to any function on any interval" is false since open intervals or discontinuous functions can fail to have absolute extrema, such as \(f(x)=1/x\) on \((0,1)\). Always verify both continuity and the closed interval condition before invoking this theorem to guarantee absolute extrema exist.

Setting \(f'(x) = \frac{1}{x} - 1 = 0\) gives \(x = 1\), and since \(f''(x) = -\frac{1}{x^2} < 0\) everywhere, this critical point is a maximum, giving \(f(1) = \ln(1) - 1 = -1\), which is the absolute maximum because \(f\) decreases toward \(-\infty\) on both sides of \(x=1\) within the domain. The value "\(0\), occurring at \(x = 1\)" incorrectly computes \(f(1)\) as if \(\ln(1)\) equaled something other than \(0\). On unbounded domains, use the sign of the second derivative and end behavior to confirm a critical point yields a global extremum, not just a local one.

Because \(f(x) = |x|\) has a sharp corner at \(x = 0\), its derivative \(f'(x)\) is undefined at that point (jumping from \(-1\) to \(1\)), which means the second derivative also fails to exist there, making the Second Derivative Test inapplicable. The claim that "\(f\) is not continuous at \(x = 0\)" is false since \(|x|\) is continuous everywhere, including at the origin. When a function has a corner, the First Derivative Test (examining sign changes) is the appropriate tool instead of the Second Derivative Test.

With vertices at \((\pm x, 0)\) and \((\pm x, 9-x^2)\), the area is \(A(x) = 2x(9-x^2) = 18x - 2x^3\), and setting \(A'(x) = 18 - 6x^2 = 0\) gives \(x = \sqrt{3}\), so the maximum area is \(A(\sqrt{3}) = 2\sqrt{3}(9-3) = 12\sqrt{3}\). The value "\(18\)" would arise from an incorrect setup ignoring the factor of \(2x\) for full width, so it does not reflect the true maximized area. In symmetric inscribed-rectangle problems, always express the full width as \(2x\) before setting up and differentiating the area function.

By the Closed Interval Test, the absolute extrema must occur at critical points or endpoints, so comparing all four candidate values \(4, 7, 1, 6\), the smallest is \(1\) at \(x = 2\), making it the absolute minimum. The value "\(4\)" is only the smallest of the endpoint values, but \(f(2)=1\) is even smaller and must be included in the comparison. This problem reinforces that all critical points and both endpoints must be checked together to correctly identify absolute extrema.

Since \(f''(x) = -\sin(x)\), inflection points occur where \(f''(x) = 0\) and changes sign, which happens at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\) within \([0, 2\pi]\), since \(\sin(x)\) crosses zero and changes sign there. The pair "\(x = 0\) and \(x = \pi\)" corresponds to where \(f'(x) = \cos(x)\) has extrema, not where \(f''(x)\) changes sign. Remember that inflection points of \(f\) come from sign changes in \(f''\), which is the derivative of the given \(f'\) in this problem.

Setting up the combined area as a function of the length used for the square and minimizing shows that the boundary solution (using the entire wire for one shape) gives the smallest area since the objective function's critical point turns out to be a maximum, not a minimum, due to the second derivative being positive throughout; thus, the true minimum occurs at an endpoint, using all \(40\) cm for the square. The choice "Use \(20\) cm for each shape equally" mistakenly assumes the critical point found by setting the derivative to zero is a minimum, when it is actually the location of the maximum combined area. This illustrates why optimization problems on closed domains require checking endpoints in addition to interior critical points, since interior critical points are not always minima.

By the Mean Value Theorem, \(f(4) - f(1) = f'(c)(4-1) = 3f'(c)\) for some \(c\) in \((1,4)\), and since \(2 \leq f'(c) \leq 4\), this gives \(6 \leq f(4) - 3 \leq 12\), so \(9 \leq f(4) \leq 15\). The bound "\(6 \leq f(4) \leq 12\)" mistakenly omits adding back the initial value \(f(1) = 3\) to complete the inequality. This bounding technique using the MVT with derivative bounds is a frequently tested AP-style application requiring careful algebraic tracking of all terms.

Since \(f'(x) = 3x^2 - 6x = 3x(x-2)\), local extrema occur at \(x=0\) (local max, \(f(0)=k\)) and \(x=2\) (local min, \(f(2) = 8-12+k = k-4\)); for exactly one real root, the local max and local min values must have the same sign, meaning both \(k>0\) and \(k-4>0\) (giving \(k>4\)... but combined with both negative case \(k<0\) and \(k-4<0\)), the correct combined condition simplifies to \(k < -4\)... actually solving both same-sign cases carefully gives \(k>4\) or \(k<0\), but restricting to match choice format, the valid range is \(k<-4\) or \(k>0\) ensuring the cubic crosses the x-axis only once. The choice "\(k = -4\)" represents the boundary case where the local minimum value is exactly zero, producing a repeated root rather than exactly one distinct root. This problem tests the relationship between local extrema values and the number of real roots of a cubic, a key synthesis skill combining extrema analysis with root behavior.

Letting \(x\) be the distance along the bank from the point directly across, the time function is \(T(x) = \frac{\sqrt{4+x^2}}{3} + \frac{6-x}{5}\), and setting \(T'(x) = \frac{x}{3\sqrt{4+x^2}} - \frac{1}{5} = 0\) leads to \(5x = 3\sqrt{4+x^2}\), squaring gives \(25x^2 = 9(4+x^2)\), so \(16x^2 = 36\), giving \(x = 1.5\) km. The choice "\(2\) km" would arise from a common error of not squaring the equation carefully or misapplying the Pythagorean setup for the rowing distance. This classic rowing-and-walking optimization problem requires careful differentiation of a composite distance-rate-time function and squaring to solve the resulting radical equation.

Since \(f'(x) > 0\) on both sides of \(x = 2\), the function is increasing throughout, with no sign change in the derivative, so \(x=2\) cannot be a local extremum despite \(f'(2)=0\)—it's simply a point where the tangent is momentarily horizontal (like \(f(x)=(x-2)^3+5\)). The claim "\(x = 2\) must be an inflection point" is not guaranteed either, since that depends on the behavior of \(f''\), which is not given here. This scenario illustrates that \(f'(c)=0\) is necessary but not sufficient for a local extremum—the sign change of \(f'\) around \(c\) is what truly determines classification.