A-Level Increasing, decreasing, stationary points Practice Questions

Free A-Level Increasing, decreasing, stationary points practice questions with full step-by-step worked solutions. Covers quadratic, turning point, minimum value, maximum value. Practise exam-style problems and check your method.

quadraticturning pointminimum valuemaximum valuenatureincreasing
A-Level70 questionsStep-by-step solutions
Question 1
3 markseasy
The curve CC has equation y=x2−4x+7y=x^{2}-4x+7. Find the coordinates of the stationary point of CC.
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Worked solution

  1. Differentiate the function

    dydx=2x−4\frac{dy}{dx}=2 x - 4

    A stationary point is where the gradient is zero, so first we need the gradient function dydx\frac{dy}{dx}. Recall the rule: multiply by the power, then subtract one from the power.

  2. Set the gradient equal to zero

    2(x−2)=02 \left(x - 2\right)=0

    At a stationary point the tangent is flat, so its gradient is 00. We set the derivative to zero and solve to find where those points are.

  3. Solve for xx

    x=2x=2

    Rearranging gives the single xx-coordinate where the curve is stationary.

  4. Find the yy-coordinate(s)

    y=3 at x=2y=3\ \text{at }x=2

    Substitute each xx back into the original curve to get the full coordinates. Always use the original yy, not the derivative.

  5. Find the second derivative

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

    To decide whether each point is a maximum or minimum we use the second derivative test, so we differentiate the gradient function once more.

  6. Apply the second derivative test

    x=2: 2>0⇒minimumx=2:\ 2>0\Rightarrow\text{minimum}

    If d2ydx2>0\frac{d^2y}{dx^2}>0 the point is a minimum (curve bends upwards); if it is <0<0 it is a maximum. Substitute each xx into the second derivative and read off the sign.

Answer
The minimum point is at (2,\ 3).
Question 2
3 markseasy
The curve y=7−4x−x2y=7-4x-x^{2} has one stationary point. State its nature.
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Worked solution

  1. Differentiate the function

    dydx=−2x−4\frac{dy}{dx}=- 2 x - 4

    A stationary point is where the gradient is zero, so first we need the gradient function dydx\frac{dy}{dx}. Recall the rule: multiply by the power, then subtract one from the power.

  2. Set the gradient equal to zero

    2(−x−2)=02 \left(- x - 2\right)=0

    At a stationary point the tangent is flat, so its gradient is 00. We set the derivative to zero and solve to find where those points are.

  3. Solve for xx

    x=−2x=-2

    Rearranging gives the single xx-coordinate where the curve is stationary.

  4. Find the yy-coordinate(s)

    y=11 at x=−2y=11\ \text{at }x=-2

    Substitute each xx back into the original curve to get the full coordinates. Always use the original yy, not the derivative.

  5. Find the second derivative

    d2ydx2=−2\frac{d^2y}{dx^2}=-2

    To decide whether each point is a maximum or minimum we use the second derivative test, so we differentiate the gradient function once more.

  6. Apply the second derivative test

    x=−2: −2<0⇒maximumx=-2:\ -2<0\Rightarrow\text{maximum}

    If d2ydx2>0\frac{d^2y}{dx^2}>0 the point is a minimum (curve bends upwards); if it is <0<0 it is a maximum. Substitute each xx into the second derivative and read off the sign.

Answer
The stationary point (−2, 11)(-2,\ 11) is a maximum.
Question 3
5 marksintermediate
Find and classify the stationary points of y=x3−3x2−9x+5y=x^{3}-3x^{2}-9x+5.
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Worked solution

  1. Differentiate the function

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

    A stationary point is where the gradient is zero, so first we need the gradient function dydx\frac{dy}{dx}. Recall the rule: multiply by the power, then subtract one from the power.

  2. Set the gradient equal to zero

    3(x2−2x−3)=03 \left(x^{2} - 2 x - 3\right)=0

    At a stationary point the tangent is flat, so its gradient is 00. We set the derivative to zero and solve to find where those points are.

  3. Factorise and solve for xx

    3(x−3)(x+1)=0 ⇒ x=−1, 33 \left(x - 3\right) \left(x + 1\right)=0\ \Rightarrow\ x=-1,\ 3

    Setting each factor to zero gives the xx-coordinate of every stationary point on the curve.

  4. Find the yy-coordinate(s)

    y=10 at x=−1,y=−22 at x=3y=10\ \text{at }x=-1,\quad y=-22\ \text{at }x=3

    Substitute each xx back into the original curve to get the full coordinates. Always use the original yy, not the derivative.

  5. Find the second derivative

    d2ydx2=6x−6\frac{d^2y}{dx^2}=6 x - 6

    To decide whether each point is a maximum or minimum we use the second derivative test, so we differentiate the gradient function once more.

  6. Apply the second derivative test

    x=−1: −12<0⇒maximum;x=3: 12>0⇒minimumx=-1:\ -12<0\Rightarrow\text{maximum};\quad x=3:\ 12>0\Rightarrow\text{minimum}

    If d2ydx2>0\frac{d^2y}{dx^2}>0 the point is a minimum (curve bends upwards); if it is <0<0 it is a maximum. Substitute each xx into the second derivative and read off the sign.

  7. State the stationary points and their nature

    maximum at (−1, 10); minimum at (3, −22)\text{maximum at }(-1,\ 10);\ \text{minimum at }(3,\ -22)

    Putting it together, we now have the exact location of each stationary point and whether it is a maximum, minimum or point of inflection.

Answer
There is a local maximum at (−1,10)(-1,10) and a local minimum at (3,−22)(3,-22).
Question 4
6 markshard
Find the coordinates of the minimum point of the curve y=x−2xy=x-2\sqrt{x}, x≥0x\geq 0.
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Worked solution

  1. Plan the method

    solve dydx=0, then test d2ydx2\text{solve }\frac{dy}{dx}=0,\ \text{then test }\frac{d^{2}y}{dx^{2}}

    Our strategy is: differentiate and set the gradient to zero to locate the stationary points, then use the second derivative to decide whether each is a maximum, minimum or point of inflection.

  2. Choose the differentiation method

    rewrite the fraction/root as a power of x\text{rewrite the fraction/root as a power of }x

    It is much easier to differentiate if we first rewrite each fraction or root as a power of xx, then use the ordinary power rule term by term.

  3. Differentiate the function

    dydx=1−1x\frac{dy}{dx}=1 - \frac{1}{\sqrt{x}}

    A stationary point is where the gradient is zero, so first we need the gradient function dydx\frac{dy}{dx}. Recall the rule: multiply by the power, then subtract one from the power.

  4. Write the gradient in a useful form

    dydx=x−1x\frac{dy}{dx}=\frac{\sqrt{x} - 1}{\sqrt{x}}

    Factorising (or tidying) the gradient function makes the next step — solving it equal to zero — much easier, because we can look at one bracket at a time.

  5. Set the gradient equal to zero

    x−1x=0\frac{\sqrt{x} - 1}{\sqrt{x}}=0

    At a stationary point the tangent is flat, so its gradient is 00. We set the derivative to zero and solve to find where those points are.

  6. Solve for xx

    x=1x=1

    Rearranging gives the single xx-coordinate where the curve is stationary.

  7. Check the domain

    x>0x>0

    The function is only defined for positive xx here, so we restrict our attention to x>0x>0.

  8. Find the yy-coordinate(s)

    y=−1 at x=1y=-1\ \text{at }x=1

    Substitute each xx back into the original curve to get the full coordinates. Always use the original yy, not the derivative.

  9. Find the second derivative

    d2ydx2=12x32\frac{d^2y}{dx^2}=\frac{1}{2 x^{\frac{3}{2}}}

    To decide whether each point is a maximum or minimum we use the second derivative test, so we differentiate the gradient function once more.

  10. Apply the second derivative test

    x=1: 12>0⇒minimumx=1:\ \frac{1}{2}>0\Rightarrow\text{minimum}

    If d2ydx2>0\frac{d^2y}{dx^2}>0 the point is a minimum (curve bends upwards); if it is <0<0 it is a maximum. Substitute each xx into the second derivative and read off the sign.

  11. State the stationary points and their nature

    minimum at (1, −1)\text{minimum at }(1,\ -1)

    Putting it together, we now have the exact location of each stationary point and whether it is a maximum, minimum or point of inflection.

Answer
The minimum point is at (1,\ -1).
Question 5
9 markschallenging
The curve has equation y=x−2sin⁡xy=x-2\sin x for 0≤x≤2π0\leq x\leq 2\pi. Find the coordinates of its minimum point, giving your answer in exact form.
Show worked solution

Worked solution

  1. State what we are looking for

    stationary points where dydx=0\text{stationary points where }\frac{dy}{dx}=0

    We want the minimum of the curve, which is one of its stationary points. So we begin by finding where the gradient is zero.

  2. Recall the derivatives we need

    ddx(x)=1,ddx(sin⁡x)=cos⁡x\frac{d}{dx}(x)=1,\quad \frac{d}{dx}(\sin x)=\cos x

    Before differentiating, remember the two standard results we need: the derivative of xx is 11, and the derivative of sin⁡x\sin x is cos⁡x\cos x. These come from the trigonometric differentiation topic.

  3. Differentiate the function

    dydx=1−2cos⁡x\frac{dy}{dx}=1-2\cos x

    Applying those rules term by term, the derivative of xx is 11 and the derivative of −2sin⁡x-2\sin x is −2cos⁡x-2\cos x.

  4. Set the gradient equal to zero

    1−2cos⁡x=01-2\cos x=0

    At a stationary point the gradient is zero, so we set the derivative to zero and solve for xx.

  5. Rearrange for cos⁡x\cos x

    cos⁡x=12\cos x=\tfrac{1}{2}

    Isolating cos⁡x\cos x gives a standard trig equation we can solve on the given interval.

  6. Recall where cosine is positive

    cos⁡x>0 in the 1st and 4th quadrants\cos x>0\text{ in the 1st and 4th quadrants}

    Because 12\tfrac{1}{2} is positive, the solutions lie in the quadrants where cosine is positive. This tells us to expect one solution near the start of the interval and one near the end.

  7. Solve on 0≤x≤2π0\leq x\leq 2\pi

    x=π3 or x=5π3x=\tfrac{\pi}{3}\ \text{or}\ x=\tfrac{5\pi}{3}

    Cosine equals 12\tfrac{1}{2} at π3\tfrac{\pi}{3} in the first quadrant and at 5π3\tfrac{5\pi}{3} in the fourth quadrant within one full turn.

  8. Find the second derivative

    d2ydx2=2sin⁡x\frac{d^{2}y}{dx^{2}}=2\sin x

    To decide which solution is the minimum we use the second derivative test, so we differentiate the gradient function again.

  9. Test x=π3x=\tfrac{\pi}{3}

    2sin⁡π3=3>02\sin\tfrac{\pi}{3}=\sqrt{3}>0

    The second derivative is positive here, so the curve bends upwards — this point is a minimum.

  10. Test x=5π3x=\tfrac{5\pi}{3}

    2sin⁡5π3=−3<02\sin\tfrac{5\pi}{3}=-\sqrt{3}<0

    The second derivative is negative here, so this point is a maximum, not the minimum we want.

  11. Select the minimum

    x=π3x=\tfrac{\pi}{3}

    Only x=π3x=\tfrac{\pi}{3} gives a minimum, so that is the xx-coordinate we use.

  12. Find the yy-coordinate

    y=π3−2sin⁡π3y=\tfrac{\pi}{3}-2\sin\tfrac{\pi}{3}

    Substitute x=π3x=\tfrac{\pi}{3} into the original equation y=x−2sin⁡xy=x-2\sin x.

  13. Evaluate exactly

    y=π3−2⋅32=π3−3y=\tfrac{\pi}{3}-2\cdot\tfrac{\sqrt{3}}{2}=\tfrac{\pi}{3}-\sqrt{3}

    Using the exact value sin⁡π3=32\sin\tfrac{\pi}{3}=\tfrac{\sqrt{3}}{2} keeps the answer exact, as required.

  14. State the coordinates

    (π3, π3−3)\left(\tfrac{\pi}{3},\ \tfrac{\pi}{3}-\sqrt{3}\right)

    Combining the results, the minimum point is at (π3, π3−3)\left(\tfrac{\pi}{3},\ \tfrac{\pi}{3}-\sqrt{3}\right).

  15. Sanity check the value

    π3≈1.05, 3≈1.73⇒y≈−0.68\tfrac{\pi}{3}\approx 1.05,\ \sqrt{3}\approx 1.73\Rightarrow y\approx -0.68

    A quick decimal check gives y≈−0.68y\approx -0.68, which is below the xx-axis and sensible for a minimum of this curve. This confirms the exact answer is reasonable.

Answer
The minimum point is at (π3, π3−3)\left(\dfrac{\pi}{3},\ \dfrac{\pi}{3}-\sqrt{3}\right).

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