AP Calculus BC Analytical Applications of Differentiation — Worked Answer Explanations

Unit 5 · 12 questions explained

1. Question 1 · Easy

   Let $f(x)=x^3-3x$ on $[-2,2]$. By the Extreme Value Theorem, the absolute maximum value of $f$ on this interval is:

   • A
     $2$Correct
   • B
     $-2$

     Why not B: Returns the absolute minimum $f(1)=-2$, confusing minimum with maximum.
   • C
     $0$

     Why not C: Returns $f(0)=0$, which is neither the maximum nor minimum.
   • D
     $8$

     Why not D: Evaluates only $x^3$ at $x=2$ ($=8$) rather than the full $f(2)=8-6=2$.
   Explanation

   Critical points: $f^{\prime }(x)=3x^2-3=0\Rightarrow x=\pm 1$. Candidate values: $f(-2)=-8+6=-2$, $f(-1)=-1+3=2$, $f(1)=1-3=-2$, $f(2)=8-6=2$. Absolute maximum is $2$, attained at $x=-1$ and $x=2$.

   Key takeaway

   To find absolute extrema on $[a,b]$: evaluate $f$ at all critical points and both endpoints; the largest value is the absolute maximum.

2. Question 2 · Easy

   For $f(x)=x^2$ on $[1,3]$, find the value $c$ guaranteed by the Mean Value Theorem.

   • A
     $c=1$

     Why not A: Returns the left endpoint; MVT guarantees $c$ in the open interval $(1,3)$.
   • B
     $c=\sqrt{3}$

     Why not B: Solves $2c=\sqrt{3}$ instead of $2c=4$ (the average rate of change is $(9-1)/(3-1)=4$).
   • C
     $c=2$Correct
   • D
     $c=3$

     Why not D: Returns the right endpoint; MVT requires $c\in (1,3)$, not at the boundary.
   Explanation

   Average rate of change: $\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4$. MVT: find $c$ with $f^{\prime }(c)=4$. $f^{\prime }(x)=2x$, so $2c=4\Rightarrow c=2\in (1,3)$.

   Key takeaway

   MVT: $f'(c) = [f(b)-f(a)]/(b-a)$; compute the average rate, then set equal to $f'(c)$ and solve.

3. Question 3 · Easy

   For $f(x)=x^3-6x^2+9x$, on what intervals is $f$ increasing?

   • A
     $(1,3)$

     Why not A: Identifies the interval where $f^{\prime }<0$ (decreasing), not where $f^{\prime }>0$ (increasing).
   • B
     $(-\infty ,1)$ and $(3,\infty )$Correct
   • C
     $(0,3)$

     Why not C: Uses $x=0$ as a boundary instead of the critical point $x=1$.
   • D
     $(-\infty ,\infty )$

     Why not D: Incorrectly concludes a cubic is always increasing.
   Explanation

   $f^{\prime }(x)=3x^2-12x+9=3(x-1)(x-3)$. Sign analysis: $f^{\prime }>0$ when $x<1$ or $x>3$. Therefore $f$ is increasing on $(-\infty ,1)$ and $(3,\infty )$.

   Key takeaway

   $f$ is increasing where $f'(x) > 0$; find critical points, draw a sign chart, and identify positive intervals.

4. Question 4 · Easy

   For $g(x)=x^4-8x^2$, use the Second Derivative Test to classify $x=0$ and $x=\pm 2$.

   • A
     $x=0$: local min; $x=\pm 2$: local max

     Why not A: Reverses the classification: $g^{\prime \prime }(0)=-16<0$ (local max) and $g^{\prime \prime }(\pm 2)=32>0$ (local min).
   • B
     $x=0$: local max; $x=\pm 2$: local minCorrect
   • C
     $x=0$: local min; $x=\pm 2$: inflection points

     Why not C: $x=\pm 2$ are critical points of $g$, not inflection points; inflection points are where $g^{\prime \prime }=0$.
   • D
     $x=0,\pm 2$: all local minima

     Why not D: Ignores the sign of $g^{\prime \prime }$; $g^{\prime \prime }(0)=-16<0$, so $x=0$ cannot be a local minimum.
   Explanation

   $g^{\prime }(x)=4x^3-16x=4x(x-2)(x+2)$; critical points $x=0,\pm 2$. $g^{\prime \prime }(x)=12x^2-16$. $g^{\prime \prime }(0)=-16<0\Rightarrow$ local max. $g^{\prime \prime }(\pm 2)=48-16=32>0\Rightarrow$ local min at both $x=2$ and $x=-2$.

   Key takeaway

   Second Derivative Test: $f''(c) > 0$ means local min; $f''(c) < 0$ means local max; $f''(c) = 0$ is inconclusive.

5. Question 5 · Easy

   Find all inflection points of $h(x)=x^4-6x^2+5$.

   • A
     $x=0$ only

     Why not A: Finds a critical point of $h^{\prime \prime }(x)=12x^2-12$; $h^{\prime \prime }(0)=-12\ne 0$, so $x=0$ is not an inflection point.
   • B
     $x=\pm 1$Correct
   • C
     $x=\pm \sqrt{3}$

     Why not C: Sets $h^{\prime }(x)=4x^3-12x=0$, finding critical points of $h$ rather than inflection points.
   • D
     $x=\pm 2$

     Why not D: Solves $12x^2-12=0$ but makes an arithmetic error writing $x^2=4$ instead of $x^2=1$.
   Explanation

   $h^{\prime }(x)=4x^3-12x$, $h^{\prime \prime }(x)=12x^2-12=12(x^2-1)$. Set $h^{\prime \prime }(x)=0$: $x=\pm 1$. Concavity check: $h^{\prime \prime }<0$ for $|x|<1$ (concave down), $h^{\prime \prime }>0$ for $|x|>1$ (concave up). Sign changes at both $x=-1$ and $x=1$: inflection points at $x=\pm 1$.

   Key takeaway

   Inflection points require both $h''(x) = 0$ AND a concavity sign change; always verify the sign of $h''$ on each side.

6. Question 6 · Easy

   A farmer has $200$ m of fencing to enclose a rectangular field against a straight river (no fence on the river side). What dimensions maximize the enclosed area?

   • A
     Width $50$ m (sides), Length $100$ m (parallel to river), Area $=5000$ m${}^2$Correct
   • B
     Width $100$ m, Length $50$ m, Area $=5000$ m${}^2$

     Why not B: Reverses width and length labels; the river-side (length) should be twice the width for optimality.
   • C
     Width $=$ Length $\approx 66.7$ m, Area $\approx 4444$ m${}^2$

     Why not C: Divides $200$ m equally into three sides instead of into two widths and one length.
   • D
     Width $=$ Length $=50$ m, Area $=2500$ m${}^2$

     Why not D: Uses all $200$ m for the two widths only ($2w=200$), ignoring the river-parallel length entirely.
   Explanation

   Let $w$ = width (perpendicular to river), $\ell$ = length (parallel to river). Constraint: $2w+\ell =200\Rightarrow \ell =200-2w$. Area: $A=w(200-2w)=200w-2w^2$. $A^{\prime }=200-4w=0\Rightarrow w=50$. Then $\ell =100$, $A=5000$ m${}^2$. $A^{\prime \prime }=-4<0$ confirms maximum.

   Key takeaway

   Set up the area as a function of one variable using the perimeter constraint (three sides), differentiate, and confirm the critical point is a maximum.

7. Question 7 · Medium

   On what interval is $f(x)=x{e}^{-x}$ concave down?

   • A
     $(2,\infty )$

     Why not A: Finds inflection at $x=2$ correctly but assigns concavity in the wrong direction; $f^{\prime \prime }<0$ to the left of $2$.
   • B
     $(-\infty ,1)$

     Why not B: Uses the local max location $x=1$ (where $f^{\prime }=0$) as the concavity boundary rather than the inflection point $x=2$.
   • C
     $(-\infty ,2)$Correct
   • D
     $(0,\infty )$

     Why not D: Correctly identifies $x=0$ as in the positive $x$-region but uses the wrong boundary; inflection is at $x=2$.
   Explanation

   $f^{\prime }(x)=e^{-x}(1-x)$. $f^{\prime \prime }(x)=-e^{-x}(1-x)+e^{-x}(-1)=e^{-x}[-(1-x)-1]=e^{-x}(x-2)$. Since $e^{-x}>0$: $f^{\prime \prime }<0\iff x-2<0\iff x<2$. So $f$ is concave down on $(-\infty ,2)$.

   Key takeaway

   Concave down: $f''(x) < 0$; compute the second derivative, factor, and identify where the sign is negative.

8. Question 8 · Medium

   A function has $f^{\prime }(x)=(x-1{)}^2(x-3)$. On which interval is $f$ strictly decreasing?

   • A
     $(1,3)$

     Why not A: For $1<x<3$: $(x-1{)}^2>0$ and $(x-3)<0$, so $f^{\prime }<0$ — this IS a decreasing interval. But the question asks for the full decreasing set.
   • B
     $(-\infty ,3)$Correct
   • C
     $(-\infty ,1)$ only

     Why not C: Identifies the left decreasing portion but misses $(1,3)$ where $f$ also decreases (the zero at $x=1$ is not a sign change).
   • D
     $(-\infty ,1)$ and $(3,\infty )$

     Why not D: Incorrectly identifies $x>3$ as decreasing; for $x>3$: $(x-1{)}^2>0$ and $(x-3)>0$, so $f^{\prime }>0$ (increasing).
   Explanation

   $f^{\prime }(x)=(x-1{)}^2(x-3)$. $(x-1{)}^2\ge 0$ everywhere, with $(x-1{)}^2=0$ only at $x=1$. So the sign of $f^{\prime }$ matches the sign of $(x-3)$: $f^{\prime }<0$ when $x<3$ (and $x\ne 1$). At $x=1$, $f^{\prime }=0$ but concavity doesn't reverse (it's an odd-multiplicity-1 factor in the sign but $(x-1{)}^2$ keeps sign positive). Therefore $f$ is decreasing on $(-\infty ,3)$.

   Key takeaway

   A squared factor $(x-a)^2 \geq 0$ never changes sign; the sign of $f'$ is controlled by the remaining factors.

9. Question 9 · Hard

   Let $f(x)=x^3+a{x}^2+bx$ have a local maximum at $x=1$ and a local minimum at $x=3$. Find $a$ and $b$.

   • A
     $a=-6$, $b=9$Correct
   • B
     $a=6$, $b=9$

     Why not B: Uses $f^{\prime }(x)=3x^2+2ax+b$ with roots at $x=1,3$ but solves Vieta's with wrong sign: sum of roots $=-2a/3$ gives $a=-6$ for sum $4$, not $a=+6$.
   • C
     $a=-6$, $b=-9$

     Why not C: Gets $a=-6$ correct but makes a sign error in the product of roots: $b/3=1\cdot 3=3$, so $b=9$, not $-9$.
   • D
     $a=-3$, $b=4$

     Why not D: Sets up equations $f^{\prime }(1)=0$ and $f^{\prime }(3)=0$ but makes arithmetic errors in the $2\times 2$ linear system.
   Explanation

   $f^{\prime }(x)=3x^2+2ax+b$. Local max at $x=1$ and local min at $x=3$ both require $f^{\prime }(x)=0$, so $f^{\prime }(x)=3(x-1)(x-3)=3x^2-12x+9$. Comparing: $2a=-12\Rightarrow a=-6$ and $b=9$. Verify: $f^{\prime \prime }(x)=6x+2a=6x-12$; $f^{\prime \prime }(1)=-6<0$ (local max ✓), $f^{\prime \prime }(3)=6>0$ (local min ✓).

   Key takeaway

   If critical points are given, factor $f'$ with those roots, then match coefficients to find the unknowns; verify with $f''$.

10. Question 10 · Hard

    A closed cylinder must hold volume $V=250\pi$ cm${}^3$. Find the radius $r$ that minimizes total surface area $S=2\pi r^2+2\pi rh$.

    • A
      $r=5$ cmCorrect
    • B
      $r=10$ cm

      Why not B: Solves $r^3=250$ using $V=500\pi$ (double the given volume), getting $r=\sqrt[3]{250}\approx 6.3$; or makes a factor-of-$2$ error.
    • C
      $r=\sqrt[3]{250}$ cm

      Why not C: Uses the volume $V=500\pi$ (twice the given $250\pi$) in the optimization equation.
    • D
      $r=\frac{5}{\sqrt[3]{2}}$ cm

      Why not D: Forgets the factor of $2$ on the top disk in $S=2\pi r^2+2\pi rh$, setting up a different equation.
    Explanation

    Constraint: $\pi r^{2}h=250\pi \Rightarrow h=250/r^2$. Substitute: $S=2\pi r^2+2\pi r(250/r^2)=2\pi r^2+500\pi /r$. $\frac{dS}{dr}=4\pi r-\frac{500\pi }{r^2}=0\Rightarrow 4r^3=500\Rightarrow r^3=125\Rightarrow r=5$ cm. $\frac{d^{2}S}{d{r}^2}=4\pi +\frac{1000\pi }{r^3}>0$ confirms minimum.

    Key takeaway

    Cylinder optimization: express $S$ in terms of $r$ alone using $h = V/(\pi r^2)$, differentiate, set to zero — gives $r^3 = V/(2\pi)$ in the standard form.

11. Question 11 · Hard

    Using the MVT on $f(x)=\sqrt{x}$ over $[4,9]$, find $c$ such that $f^{\prime }(c)$ equals the average rate of change.

    • A
      $c=\frac{25}{4}$Correct
    • B
      $c=\frac{13}{2}$

      Why not B: Takes the arithmetic midpoint $(4+9)/2=6.5$ of the interval instead of solving the MVT equation.
    • C
      $c=6$

      Why not C: Approximates $\sqrt{c}=5$ (midpoint of $\sqrt{4}=2$ and $\sqrt{9}=3$) but then takes $c=5^2/\sqrt{5}$ incorrectly.
    • D
      $c=\sqrt{\frac{25}{4}}$

      Why not D: Takes the square root of $c=25/4$ rather than returning the $c$ value itself; $\sqrt{25/4}=5/2\notin (4,9)$.
    Explanation

    Average rate: $\frac{\sqrt{9}-\sqrt{4}}{9-4}=\frac{3-2}{5}=\frac{1}{5}$. Set $f^{\prime }(c)=\frac{1}{2\sqrt{c}}=\frac{1}{5}\Rightarrow 2\sqrt{c}=5\Rightarrow \sqrt{c}=\frac{5}{2}\Rightarrow c=\frac{25}{4}=6.25$. Since $4<6.25<9$: $c=25/4$ ✓.

    Key takeaway

    MVT on $\sqrt{x}$: set $1/(2\sqrt{c})$ equal to the average rate and solve for $c$; verify $c$ lies in the open interval.

12. Question 12 · Hard

    Let $f$ be twice differentiable with $f^{\prime }(2)=0$, $f^{\prime \prime }(2)=0$, and $f^{\prime \prime \prime }(2)\ne 0$. What can be concluded about $x=2$?

    • A
      $x=2$ is a local minimum because $f^{\prime }(2)=0$.

      Why not A: The second derivative test is inconclusive when $f^{\prime \prime }(2)=0$; cannot conclude minimum from $f^{\prime }(2)=0$ alone.
    • B
      $x=2$ is an inflection point with a horizontal tangent.Correct
    • C
      $x=2$ is a saddle point but not an inflection point.

      Why not C: 'Saddle point' is not a standard single-variable calculus term for a horizontal tangent inflection; the correct term is inflection point with $f^{\prime }(2)=0$.
    • D
      No conclusion can be drawn without higher derivatives.

      Why not D: The given information ($f^{\prime \prime \prime }(2)\ne 0$) is sufficient to conclude the point is an inflection: concavity changes across $x=2$.
    Explanation

    $f^{\prime }(2)=0$ means a horizontal tangent. $f^{\prime \prime }(2)=0$ means the second derivative test is inconclusive. $f^{\prime \prime \prime }(2)\ne 0$ means $f^{\prime \prime }$ changes sign at $x=2$ (since $f^{\prime \prime }$ has an odd-order root there). Therefore concavity changes at $x=2$: it is an inflection point. Since $f^{\prime }(2)=0$, it is an inflection point with a horizontal tangent (sometimes called a 'stationary inflection point' or 'saddle point' in informal usage).

    Key takeaway

    When $f'(c) = 0$, $f''(c) = 0$, and $f'''(c) \neq 0$: $f''$ has a simple root at $c$, concavity changes, so $c$ is an inflection point (with a horizontal tangent).