AP Calculus AB Analytical Applications of Differentiation — Worked Answer Explanations

Unit 5 · 18% of the AP exam · 12 questions explained

1. Question 1 · Easy

   If $f^{\prime }(x)>0$ on $(a,b)$, then $f$ is:

   • A
     Concave up on $(a,b)$.

     Why not A: Concavity depends on the sign of $f^{\prime \prime }$, not $f^{\prime }$.
   • B
     Increasing on $(a,b)$.Correct
   • C
     Decreasing on $(a,b)$.

     Why not C: Decreasing corresponds to $f^{\prime }(x)<0$.
   • D
     Constant on $(a,b)$.

     Why not D: Constant functions have $f^{\prime }(x)=0$, not positive.
   Explanation

   By the First Derivative Test for monotonicity, $f^{\prime }>0$ on an interval means $f$ is increasing there — each step right, $f$ goes up.

   Key takeaway

   The sign of $f'$ determines monotonicity: positive means increasing, negative means decreasing.

2. Question 2 · Easy

   The function $f$ is continuous on $[2,8]$, differentiable on $(2,8)$, with $f(2)=1$ and $f(8)=13$. Which theorem guarantees a $c\in (2,8)$ where $f^{\prime }(c)=2$?

   • A
     Extreme Value Theorem

     Why not A: EVT guarantees max/min exist, not a specific slope.
   • B
     Intermediate Value Theorem

     Why not B: IVT applies to function values, not derivative values.
   • C
     Mean Value TheoremCorrect
   • D
     Rolle's Theorem

     Why not D: Rolle's requires $f(a)=f(b)$; here $f(2)\ne f(8)$.
   Explanation

   The MVT states that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in (a,b)$ with $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}=\frac{13-1}{8-2}=\frac{12}{6}=2$.

   Key takeaway

   The Mean Value Theorem guarantees an interior point where the instantaneous rate equals the average rate over $[a,b]$.

3. Question 3 · Easy

   If $f^{\prime \prime }(x)<0$ on an interval, then the graph of $f$ is:

   • A
     Concave up.

     Why not A: Concave up corresponds to $f^{\prime \prime }>0$.
   • B
     Decreasing.

     Why not B: Decreasing depends on $f^{\prime }<0$, not $f^{\prime \prime }$.
   • C
     Concave down.Correct
   • D
     At a local maximum.

     Why not D: $f^{\prime \prime }<0$ alone does not locate a maximum; $f^{\prime }=0$ is also required.
   Explanation

   The sign of $f^{\prime \prime }$ determines concavity. $f^{\prime \prime }<0$ means $f^{\prime }$ is decreasing, so the graph bends downward — concave down.

   Key takeaway

   $f'' < 0$ means concave down (frowning shape); $f'' > 0$ means concave up (smiling shape).

4. Question 4 · Easy

   A function $f$ is continuous on $[-3,5]$. Which theorem guarantees that $f$ attains both a maximum and a minimum value on this interval?

   • A
     Mean Value Theorem

     Why not A: MVT is about derivatives, not extreme values.
   • B
     Intermediate Value Theorem

     Why not B: IVT guarantees a function value between two outputs, not an extreme value.
   • C
     Rolle's Theorem

     Why not C: Rolle's requires equal endpoint values and concludes $f^{\prime }(c)=0$.
   • D
     Extreme Value TheoremCorrect
   Explanation

   The Extreme Value Theorem (EVT) states that a function continuous on a closed bounded interval $[a,b]$ must attain an absolute maximum and an absolute minimum on that interval.

   Key takeaway

   EVT guarantees absolute extrema exist on closed intervals for continuous functions.

5. Question 5 · Medium

   Let $f(x)=x^3-3x^2-9x+5$. On what interval(s) is $f$ increasing?

   • A
     $(-\infty ,-1)$ and $(3,\infty )$

     Why not A: Confused the critical points: $x=-1$ is a local max and $x=3$ is a local min.
   • B
     $(-1,3)$

     Why not B: This is where $f$ is decreasing, not increasing.
   • C
     $(-\infty ,-1)$ and $(3,\infty )$Correct
   • D
     $(-\infty ,3)$

     Why not D: Did not check the sign of $f^{\prime }$ correctly between critical points.
   Explanation

   $f^{\prime }(x)=3x^2-6x-9=3(x^2-2x-3)=3(x-3)(x+1)$. Critical points at $x=-1$ and $x=3$. Sign chart: $f^{\prime }>0$ for $x<-1$ and $x>3$; $f^{\prime }<0$ for $-1<x<3$. So $f$ is increasing on $(-\infty ,-1)$ and $(3,\infty )$.

   Key takeaway

   Find where $f' > 0$ using a sign chart after factoring; increasing intervals come from positive regions of $f'$.

6. Question 6 · Medium

   The graph of $f^{\prime }$ is shown (not provided here — described): $f^{\prime }(x)>0$ for $x<2$, $f^{\prime }(2)=0$, and $f^{\prime }(x)<0$ for $x>2$. What can be concluded about $f$ at $x=2$?

   • A
     $f$ has a local minimum at $x=2$.

     Why not A: A local min requires $f^{\prime }$ changing from negative to positive, not positive to negative.
   • B
     $f$ has a local maximum at $x=2$.Correct
   • C
     $f$ has an inflection point at $x=2$.

     Why not C: Inflection points require a sign change in $f^{\prime \prime }$, not just $f^{\prime }=0$.
   • D
     $f$ is not differentiable at $x=2$.

     Why not D: $f^{\prime }(2)=0$ means $f$ is differentiable there.
   Explanation

   First Derivative Test: if $f^{\prime }$ changes from positive to negative at $x=2$, then $f$ has a local maximum at $x=2$. This matches the described behavior of $f^{\prime }$.

   Key takeaway

   First Derivative Test: $f'$ changes $+ \to -$ at a critical point indicates a local maximum.

7. Question 7 · Medium

   Find all inflection points of $f(x)=x^4-4x^3$.

   • A
     $x=0$ only

     Why not A: $x=0$ is an inflection point too — sign of $f^{\prime \prime }$ changes there as well.
   • B
     $x=3$ only

     Why not B: Used $f^{\prime }(x)=0$ (critical point of $f$) rather than $f^{\prime \prime }(x)=0$.
   • C
     $x=2$ only

     Why not C: Found one inflection point but missed $x=0$ where $f^{\prime \prime }$ also changes sign.
   • D
     $x=0$ and $x=2$Correct
   Explanation

   $f^{\prime }(x)=4x^3-12x^2$. $f^{\prime \prime }(x)=12x^2-24x=12x(x-2)$. $f^{\prime \prime }=0$ at $x=0$ and $x=2$. Sign check: $f^{\prime \prime }>0$ for $x<0$; $f^{\prime \prime }<0$ for $0<x<2$; $f^{\prime \prime }>0$ for $x>2$. At $x=0$: $f^{\prime \prime }$ changes $+\to -$ — inflection point. At $x=2$: $f^{\prime \prime }$ changes $-\to +$ — inflection point. Both $x=0$ and $x=2$ are inflection points.

   Key takeaway

   Inflection points require $f'' = 0$ AND a sign change in $f''$; check all candidates by examining sign changes.

8. Question 8 · Medium

   Find the absolute maximum of $f(x)=-x^2+4x+1$ on the interval $[0,5]$.

   • A
     $1$

     Why not A: Only evaluated $f(0)$, the left endpoint.
   • B
     $5$Correct
   • C
     $4$

     Why not C: Found the $x$-value of the critical point, not the function value.
   • D
     $6$

     Why not D: Evaluated incorrectly at the critical point: $f(2)=-4+8+1=5$, not $6$.
   Explanation

   Critical point: $f^{\prime }(x)=-2x+4=0\Rightarrow x=2$. Evaluate candidates: $f(0)=1$, $f(2)=-4+8+1=5$, $f(5)=-25+20+1=-4$. The absolute maximum is $f(2)=5$.

   Key takeaway

   Closed Interval Method: evaluate $f$ at all critical points in $(a,b)$ and at both endpoints; the largest value is the absolute max.

9. Question 9 · Medium

   Using the Second Derivative Test, classify the critical point $x=1$ of $f(x)=x^3-3x+2$.

   • A
     Local maximum

     Why not A: Misidentified the sign of $f^{\prime \prime }(1)$.
   • B
     Local minimumCorrect
   • C
     Inflection point

     Why not C: $f^{\prime \prime }(1)\ne 0$, so the Second Derivative Test is conclusive.
   • D
     $x=1$ is not a critical point.

     Why not D: $f^{\prime }(1)=3(1{)}^2-3=0$, confirming it is a critical point.
   Explanation

   $f^{\prime }(x)=3x^2-3$, $f^{\prime }(1)=0$ ✓. $f^{\prime \prime }(x)=6x$, $f^{\prime \prime }(1)=6>0$. Since $f^{\prime \prime }(1)>0$, the function is concave up at $x=1$, so this is a local minimum.

   Key takeaway

   Second Derivative Test: $f'(c)=0$ and $f''(c)>0$ indicates a local minimum; $f''(c)<0$ indicates a local maximum.

10. Question 10 · Hard

    A farmer wants to enclose a rectangular plot next to a river (no fence needed along the river) using $200$ m of fencing. What dimensions maximize the enclosed area?

    • A
      $50$ m $\times$ $100$ mCorrect
    • B
      $100$ m $\times$ $100$ m

      Why not B: Ignored the constraint that only 200 m of fencing is available.
    • C
      $50$ m $\times$ $50$ m

      Why not C: Used all four sides in the constraint instead of three.
    • D
      $66.7$ m $\times$ $66.7$ m

      Why not D: Divided the perimeter equally among three sides incorrectly.
    Explanation

    Let $x$ = width (two sides) and $y$ = length (parallel to river). Constraint: $2x+y=200$, so $y=200-2x$. Area: $A=xy=x(200-2x)=200x-2x^2$. Maximize: $A^{\prime }(x)=200-4x=0\Rightarrow x=50$. Then $y=200-100=100$. Maximum area is $50\times 100=5000$ m².

    Key takeaway

    Optimization: write objective in one variable using the constraint, differentiate, set equal to zero, verify it's a max.

11. Question 11 · Hard

    A function $f$ has $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)=0$. What can be concluded?

    • A
      $f$ has a local maximum at $x=c$.

      Why not A: Cannot conclude max from $f^{\prime \prime }(c)=0$ alone.
    • B
      $f$ has a local minimum at $x=c$.

      Why not B: Cannot conclude min from $f^{\prime \prime }(c)=0$ alone.
    • C
      $f$ has an inflection point at $x=c$.

      Why not C: Inflection requires a sign change in $f^{\prime \prime }$, not just $f^{\prime \prime }(c)=0$.
    • D
      The Second Derivative Test is inconclusive; further analysis is needed.Correct
    Explanation

    When both $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)=0$, the Second Derivative Test gives no information. The critical point could be a local max, local min, or inflection point. For example, $f(x)=x^4$ has $f^{\prime }(0)=f^{\prime \prime }(0)=0$ but $x=0$ is a local minimum; $f(x)=x^3$ has the same but $x=0$ is an inflection point.

    Key takeaway

    $f''(c) = 0$ at a critical point makes the Second Derivative Test inconclusive; use a sign chart of $f'$ instead.

12. Question 12 · Hard

    A particle moves so that $x(t)=t^3-6t^2+9t$ for $t\ge 0$. On which interval(s) is the particle's speed increasing?

    • A
      $(0,1)$ only

      Why not A: On $(0,1)$: $v>0$ but $a<0$, so speed is decreasing, not increasing.
    • B
      $(1,2)$ and $(3,\infty )$Correct
    • C
      $(3,\infty )$ only

      Why not C: Missed the interval $(1,2)$ where both $v<0$ and $a<0$.
    • D
      $(0,3)$

      Why not D: Did not check whether $v$ and $a$ have the same sign throughout $(0,3)$.
    Explanation

    Speed increases when $v$ and $a$ have the same sign. $v(t)=3t^2-12t+9=3(t-1)(t-3)$: positive on $(0,1)$ and $(3,\infty )$, negative on $(1,3)$. $a(t)=6t-12=6(t-2)$: negative on $(0,2)$, positive on $(2,\infty )$. Same sign: $v<0$ and $a<0$ on $(1,2)$; $v>0$ and $a>0$ on $(3,\infty )$. Speed is increasing on $(1,2)$ and $(3,\infty )$.

    Key takeaway

    Speed increases when velocity and acceleration have the same sign; construct separate sign charts for $v(t)$ and $a(t)$.