A-Level Second derivatives and curve behaviour Practice Questions

Free A-Level Second derivatives and curve behaviour practice questions with full step-by-step worked solutions. Covers second derivative, concavity. Practise exam-style problems and check your method.

second derivativeconcavity
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
The curve has equation y=x2−4x+7y=x^{2} - 4 x + 7. Find d2ydx2\frac{d^2y}{dx^2}.
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Worked solution

  1. Differentiate to find the first derivative

    dydx=2x−4\frac{dy}{dx}=2 x - 4

    Differentiate the function term by term.

  2. Differentiate the first derivative again

    ddx(2x−4)=2\frac{d}{dx}\left(2 x - 4\right)=2

    Differentiating the gradient function gives the second derivative.

  3. State the second derivative

    d2ydx2=2\frac{d^2y}{dx^2}=2

    This is the required second derivative.

Answer
22
Question 2
2 markseasy
For the curve y=f(x)y=f(x), a point of inflection occurs:
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Worked solution

  1. Recall the necessary condition for an inflection point

    d2ydx2=0\frac{d^2y}{dx^2}=0

    The second derivative is zero at a candidate inflection point.

  2. Add the concavity requirement

    d2ydx2 changes sign\frac{d^2y}{dx^2}\ \text{changes sign}

    The concavity must actually change for a point of inflection.

  3. State the condition for a point of inflection

    d2ydx2=0 and it changes sign\frac{d^2y}{dx^2}=0\ \text{and it changes sign}

    The second derivative must be zero and change sign for an inflection point.

Answer
Where f''(x)=0 and the concavity changes
Question 3
3 marksintermediate
Determine whether the curve y=x3−6x2+9x+1y=x^{3} - 6 x^{2} + 9 x + 1 is convex or concave at x=3x=3.
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Worked solution

  1. Write down the equation of the curve

    y=x3−6x2+9x+1y=x^{3} - 6 x^{2} + 9 x + 1

    We will study its gradient and concavity.

  2. Differentiate to find the first derivative

    dydx=3x2−12x+9\frac{dy}{dx}=3 x^{2} - 12 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

    3x2−12x+9=0 ⇒ x=1, x=33 x^{2} - 12 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=5y=5

    Substitute the x-value into the original equation.

  6. State whether the curve is convex or concave

    d2ydx2∣x=3=6>0 ⇒ convex\left.\frac{d^2y}{dx^2}\right|_{x=3}=6>0\ \Rightarrow\ \text{convex}

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

Answer
Convex (the curve bends upwards)
Question 4
5 markshard
Determine whether the curve y=x4−2x2y=x^{4} - 2 x^{2} is convex or concave at x=0x=0.
Show worked solution

Worked solution

  1. Write down the equation of the curve

    y=x4−2x2y=x^{4} - 2 x^{2}

    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=−1y=-1

    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=1

    y=−1y=-1

    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. State whether the curve is convex or concave

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

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