Hard A-Level Second derivatives and curve behaviour Questions

Challenging, exam-style A-Level Second derivatives and curve behaviour questions with worked solutions. Stretch yourself on the hardest second derivative, concavity problems.

second derivativeconcavity
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
Determine whether the curve y=x4−8x2y=x^{4} - 8 x^{2} is convex or concave at x=2x=2.
Show worked solution

Worked solution

  1. Write down the equation of the curve

    y=x4−8x2y=x^{4} - 8 x^{2}

    We will study its gradient and concavity.

  2. Differentiate to find the first derivative

    dydx=4x3−16x\frac{dy}{dx}=4 x^{3} - 16 x

    Differentiate term by term using the power rule.

  3. Factorise the first derivative

    dydx=4x(x−2)(x+2)\frac{dy}{dx}=4 x \left(x - 2\right) \left(x + 2\right)

    Factorising makes the stationary points easy to find.

  4. Solve the first derivative equal to zero

    4x3−16x=0 ⇒ x=−2, x=0, x=24 x^{3} - 16 x=0\ \Rightarrow\ x=-2,\ x=0,\ x=2

    Stationary points occur where the gradient is zero.

  5. Find the y-coordinate at x=-2

    y=−16y=-16

    Substitute the x-value into the original equation.

  6. Find the y-coordinate at x=0

    y=0y=0

    Substitute the x-value into the original equation.

  7. Find the y-coordinate at x=2

    y=−16y=-16

    Substitute the x-value into the original equation.

  8. Differentiate again to find the second derivative

    d2ydx2=4(3x2−4)\frac{d^2y}{dx^2}=4 \left(3 x^{2} - 4\right)

    The second derivative measures how the gradient is changing.

  9. Evaluate the second derivative at x=-2

    d2ydx2∣x=−2=32\left.\frac{d^2y}{dx^2}\right|_{x=-2}=32

    The sign of the second derivative classifies the stationary point.

  10. Classify the stationary point at x=-2

    32>0 ⇒ minimum32>0\ \Rightarrow\ \text{minimum}

    A positive second derivative means a minimum.

  11. Evaluate the second derivative at x=0

    d2ydx2∣x=0=−16\left.\frac{d^2y}{dx^2}\right|_{x=0}=-16

    The sign of the second derivative classifies the stationary point.

  12. Classify the stationary point at x=0

    −16<0 ⇒ maximum-16<0\ \Rightarrow\ \text{maximum}

    A negative second derivative means a maximum.

  13. Evaluate the second derivative at x=2

    d2ydx2∣x=2=32\left.\frac{d^2y}{dx^2}\right|_{x=2}=32

    The sign of the second derivative classifies the stationary point.

  14. Classify the stationary point at x=2

    32>0 ⇒ minimum32>0\ \Rightarrow\ \text{minimum}

    A positive second derivative means a minimum.

  15. State whether the curve is convex or concave

    d2ydx2∣x=2=32>0 ⇒ convex\left.\frac{d^2y}{dx^2}\right|_{x=2}=32>0\ \Rightarrow\ \text{convex}

    A positive second derivative means convex; a negative one means concave.

Answer
Convex (the curve bends upwards)
Question 2
8 markschallenging
The curve y=x3−3x2−9x+5y=x^{3} - 3 x^{2} - 9 x + 5 has a stationary point at x=−1x=-1. Use the second derivative test to classify this stationary point.
Show worked solution

Worked solution

  1. Write down the equation of the curve

    y=x3−3x2−9x+5y=x^{3} - 3 x^{2} - 9 x + 5

    We will study its gradient and concavity.

  2. Differentiate to find the first derivative

    dydx=3x2−6x−9\frac{dy}{dx}=3 x^{2} - 6 x - 9

    Differentiate term by term using the power rule.

  3. Factorise the first derivative

    dydx=3(x−3)(x+1)\frac{dy}{dx}=3 \left(x - 3\right) \left(x + 1\right)

    Factorising makes the stationary points easy to find.

  4. Solve the first derivative equal to zero

    3x2−6x−9=0 ⇒ x=−1, x=33 x^{2} - 6 x - 9=0\ \Rightarrow\ x=-1,\ x=3

    Stationary points occur where the gradient is zero.

  5. Find the y-coordinate at x=-1

    y=10y=10

    Substitute the x-value into the original equation.

  6. Find the y-coordinate at x=3

    y=−22y=-22

    Substitute the x-value into the original equation.

  7. Differentiate again to find the second derivative

    d2ydx2=6(x−1)\frac{d^2y}{dx^2}=6 \left(x - 1\right)

    The second derivative measures how the gradient is changing.

  8. Evaluate the second derivative at x=-1

    d2ydx2∣x=−1=−12\left.\frac{d^2y}{dx^2}\right|_{x=-1}=-12

    The sign of the second derivative classifies the stationary point.

  9. Classify the stationary point at x=-1

    −12<0 ⇒ maximum-12<0\ \Rightarrow\ \text{maximum}

    A negative second derivative means a maximum.

  10. Evaluate the second derivative at x=3

    d2ydx2∣x=3=12\left.\frac{d^2y}{dx^2}\right|_{x=3}=12

    The sign of the second derivative classifies the stationary point.

  11. Classify the stationary point at x=3

    12>0 ⇒ minimum12>0\ \Rightarrow\ \text{minimum}

    A positive second derivative means a minimum.

  12. Solve the second derivative equal to zero

    6(x−1)=0 ⇒ x=16 \left(x - 1\right)=0\ \Rightarrow\ x=1

    Points of inflection can only occur where the second derivative is zero.

  13. Confirm the concavity changes at x=1

    d2ydx2 changes sign at x=1\frac{d^2y}{dx^2}\text{ changes sign at }x=1

    A point of inflection requires a genuine change of concavity.

  14. Find the y-coordinate of the inflection point at x=1

    y=−6y=-6

    Substitute into the original equation for the full coordinate.

  15. Apply the second derivative test

    d2ydx2∣x=−1=−12<0 ⇒ maximum\left.\frac{d^2y}{dx^2}\right|_{x=-1}=-12<0\ \Rightarrow\ \text{maximum}

    The sign of the second derivative gives the nature of the point.

Answer
Maximum
Question 3
8 markschallenging
The curve y=x4−2x2+3y=x^{4} - 2 x^{2} + 3 has a stationary point at x=0x=0. Use the second derivative test to classify this stationary point.
Show worked solution

Worked solution

  1. Write down the equation of the curve

    y=x4−2x2+3y=x^{4} - 2 x^{2} + 3

    We will study its gradient and concavity.

  2. Differentiate to find the first derivative

    dydx=4x3−4x\frac{dy}{dx}=4 x^{3} - 4 x

    Differentiate term by term using the power rule.

  3. Factorise the first derivative

    dydx=4x(x−1)(x+1)\frac{dy}{dx}=4 x \left(x - 1\right) \left(x + 1\right)

    Factorising makes the stationary points easy to find.

  4. Solve the first derivative equal to zero

    4x3−4x=0 ⇒ x=−1, x=0, x=14 x^{3} - 4 x=0\ \Rightarrow\ x=-1,\ x=0,\ x=1

    Stationary points occur where the gradient is zero.

  5. Find the y-coordinate at x=-1

    y=2y=2

    Substitute the x-value into the original equation.

  6. Find the y-coordinate at x=0

    y=3y=3

    Substitute the x-value into the original equation.

  7. Find the y-coordinate at x=1

    y=2y=2

    Substitute the x-value into the original equation.

  8. Differentiate again to find the second derivative

    d2ydx2=4(3x2−1)\frac{d^2y}{dx^2}=4 \left(3 x^{2} - 1\right)

    The second derivative measures how the gradient is changing.

  9. Evaluate the second derivative at x=-1

    d2ydx2∣x=−1=8\left.\frac{d^2y}{dx^2}\right|_{x=-1}=8

    The sign of the second derivative classifies the stationary point.

  10. Classify the stationary point at x=-1

    8>0 ⇒ minimum8>0\ \Rightarrow\ \text{minimum}

    A positive second derivative means a minimum.

  11. Evaluate the second derivative at x=0

    d2ydx2∣x=0=−4\left.\frac{d^2y}{dx^2}\right|_{x=0}=-4

    The sign of the second derivative classifies the stationary point.

  12. Classify the stationary point at x=0

    −4<0 ⇒ maximum-4<0\ \Rightarrow\ \text{maximum}

    A negative second derivative means a maximum.

  13. Evaluate the second derivative at x=1

    d2ydx2∣x=1=8\left.\frac{d^2y}{dx^2}\right|_{x=1}=8

    The sign of the second derivative classifies the stationary point.

  14. Classify the stationary point at x=1

    8>0 ⇒ minimum8>0\ \Rightarrow\ \text{minimum}

    A positive second derivative means a minimum.

  15. Apply the second derivative test

    d2ydx2∣x=0=−4<0 ⇒ maximum\left.\frac{d^2y}{dx^2}\right|_{x=0}=-4<0\ \Rightarrow\ \text{maximum}

    The sign of the second derivative gives the nature of the point.

Answer
Maximum
Question 4
8 markschallenging
Determine whether the curve y=x4−18x2+5y=x^{4} - 18 x^{2} + 5 is convex or concave at x=0x=0.
Show worked solution

Worked solution

  1. Write down the equation of the curve

    y=x4−18x2+5y=x^{4} - 18 x^{2} + 5

    We will study its gradient and concavity.

  2. Differentiate to find the first derivative

    dydx=4x3−36x\frac{dy}{dx}=4 x^{3} - 36 x

    Differentiate term by term using the power rule.

  3. Factorise the first derivative

    dydx=4x(x−3)(x+3)\frac{dy}{dx}=4 x \left(x - 3\right) \left(x + 3\right)

    Factorising makes the stationary points easy to find.

  4. Solve the first derivative equal to zero

    4x3−36x=0 ⇒ x=−3, x=0, x=34 x^{3} - 36 x=0\ \Rightarrow\ x=-3,\ x=0,\ x=3

    Stationary points occur where the gradient is zero.

  5. Find the y-coordinate at x=-3

    y=−76y=-76

    Substitute the x-value into the original equation.

  6. Find the y-coordinate at x=0

    y=5y=5

    Substitute the x-value into the original equation.

  7. Find the y-coordinate at x=3

    y=−76y=-76

    Substitute the x-value into the original equation.

  8. Differentiate again to find the second derivative

    d2ydx2=12(x2−3)\frac{d^2y}{dx^2}=12 \left(x^{2} - 3\right)

    The second derivative measures how the gradient is changing.

  9. Evaluate the second derivative at x=-3

    d2ydx2∣x=−3=72\left.\frac{d^2y}{dx^2}\right|_{x=-3}=72

    The sign of the second derivative classifies the stationary point.

  10. Classify the stationary point at x=-3

    72>0 ⇒ minimum72>0\ \Rightarrow\ \text{minimum}

    A positive second derivative means a minimum.

  11. Evaluate the second derivative at x=0

    d2ydx2∣x=0=−36\left.\frac{d^2y}{dx^2}\right|_{x=0}=-36

    The sign of the second derivative classifies the stationary point.

  12. Classify the stationary point at x=0

    −36<0 ⇒ maximum-36<0\ \Rightarrow\ \text{maximum}

    A negative second derivative means a maximum.

  13. Evaluate the second derivative at x=3

    d2ydx2∣x=3=72\left.\frac{d^2y}{dx^2}\right|_{x=3}=72

    The sign of the second derivative classifies the stationary point.

  14. Classify the stationary point at x=3

    72>0 ⇒ minimum72>0\ \Rightarrow\ \text{minimum}

    A positive second derivative means a minimum.

  15. State whether the curve is convex or concave

    d2ydx2∣x=0=−36<0 ⇒ concave\left.\frac{d^2y}{dx^2}\right|_{x=0}=-36<0\ \Rightarrow\ \text{concave}

    A positive second derivative means convex; a negative one means concave.

Answer
Concave (the curve bends downwards)
Question 5
8 markschallenging
The curve y=x4−8x2y=x^{4} - 8 x^{2} has a stationary point at x=−2x=-2. Use the second derivative test to classify this stationary point.
Show worked solution

Worked solution

  1. Write down the equation of the curve

    y=x4−8x2y=x^{4} - 8 x^{2}

    We will study its gradient and concavity.

  2. Differentiate to find the first derivative

    dydx=4x3−16x\frac{dy}{dx}=4 x^{3} - 16 x

    Differentiate term by term using the power rule.

  3. Factorise the first derivative

    dydx=4x(x−2)(x+2)\frac{dy}{dx}=4 x \left(x - 2\right) \left(x + 2\right)

    Factorising makes the stationary points easy to find.

  4. Solve the first derivative equal to zero

    4x3−16x=0 ⇒ x=−2, x=0, x=24 x^{3} - 16 x=0\ \Rightarrow\ x=-2,\ x=0,\ x=2

    Stationary points occur where the gradient is zero.

  5. Find the y-coordinate at x=-2

    y=−16y=-16

    Substitute the x-value into the original equation.

  6. Find the y-coordinate at x=0

    y=0y=0

    Substitute the x-value into the original equation.

  7. Find the y-coordinate at x=2

    y=−16y=-16

    Substitute the x-value into the original equation.

  8. Differentiate again to find the second derivative

    d2ydx2=4(3x2−4)\frac{d^2y}{dx^2}=4 \left(3 x^{2} - 4\right)

    The second derivative measures how the gradient is changing.

  9. Evaluate the second derivative at x=-2

    d2ydx2∣x=−2=32\left.\frac{d^2y}{dx^2}\right|_{x=-2}=32

    The sign of the second derivative classifies the stationary point.

  10. Classify the stationary point at x=-2

    32>0 ⇒ minimum32>0\ \Rightarrow\ \text{minimum}

    A positive second derivative means a minimum.

  11. Evaluate the second derivative at x=0

    d2ydx2∣x=0=−16\left.\frac{d^2y}{dx^2}\right|_{x=0}=-16

    The sign of the second derivative classifies the stationary point.

  12. Classify the stationary point at x=0

    −16<0 ⇒ maximum-16<0\ \Rightarrow\ \text{maximum}

    A negative second derivative means a maximum.

  13. Evaluate the second derivative at x=2

    d2ydx2∣x=2=32\left.\frac{d^2y}{dx^2}\right|_{x=2}=32

    The sign of the second derivative classifies the stationary point.

  14. Classify the stationary point at x=2

    32>0 ⇒ minimum32>0\ \Rightarrow\ \text{minimum}

    A positive second derivative means a minimum.

  15. Apply the second derivative test

    d2ydx2∣x=−2=32>0 ⇒ minimum\left.\frac{d^2y}{dx^2}\right|_{x=-2}=32>0\ \Rightarrow\ \text{minimum}

    The sign of the second derivative gives the nature of the point.

Answer
Minimum

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