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Calculus (JAMB bridge) · Lesson 3 of 5

Maxima, minima and rates of change

Stationary points where dy/dx = 0, telling a maximum from a minimum, maximum and minimum values, problems about the largest or smallest value, rates of change, and velocity and acceleration.

17 minYou should already know: Quadratics & their graphs Coordinate geometry
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Stationary points

At the top of a hill or the bottom of a valley, the tangent is flat: dydx=0\dfrac{dy}{dx} = 0. These points are called stationary points. To find them, differentiate, set dydx=0\dfrac{dy}{dx} = 0, and solve.

To tell which is which, look at the second derivative, or at how the gradient’s sign changes:

  • d2ydx2<0\dfrac{d^2y}{dx^2} < 0: a maximum (the gradient goes +,0,−+, 0, -);
  • d2ydx2>0\dfrac{d^2y}{dx^2} > 0: a minimum (the gradient goes −,0,+-, 0, +).
xmaxmin+−−+
Maximum and minimumFlat tangents; the gradient's sign changes + 0 − at a maximum and − 0 + at a minimum

Try it

Maximum and minimum pointsSlide x along the curve
1234−22468xy+−+00
9dy/dx−12d²y/dx²risingpoint
dy/dx = 3x² − 12x + 9 = 9: the curve is rising here. Stationary points are where dy/dx = 0; for this curve that is x = 1 and x = 3. Slide onto one.

Slide xx along each curve. The strip underneath shows where the gradient is positive and where it is negative. The stationary points are where it changes.

Worked example · JAMB 1992

JAMB 1992 · UME · Q39

Obtain the maximum value of the function f(x)=x3−12x+11f(x) = x^3 - 12x + 11.

  1. Stationary points

    • f′(x)=3x2−12=0{f'(x) = 3x^2 - 12 = 0}, so x2=4{x^2 = 4}.
    • x=2{x = 2} or x=−2{x = -2}.

    Think first. Solve f′(x) = 0.

  2. Which is the maximum?

    f′′(−2)=−12<0{f''(-2) = -12 < 0}, so x=−2{x = -2} gives the maximum.

    Think first. f″(x) = 6x. Where is it negative?

  3. The value

    f(−2)=−8+24+11=27{f(-2) = -8 + 24 + 11 = 27}: option D.

    Think first. Work out f(−2).

Largest and smallest

For a problem about the greatest or least value:

  1. write the quantity as a function of one variable;
  2. differentiate, set the derivative to 0 and solve;
  3. check it is a maximum or minimum with the second derivative.

Rates of change

dVdr\dfrac{dV}{dr} is the rate at which VV changes as rr changes. When two quantities both change with time, link their rates with the chain rule:

dAdt=dAdr×drdt\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}
dA/dt = dA/dr × dr/dt
from the formula × the rate you are given
Connected ratesThe derivative from the formula, times the rate you are given

Worked example · JAMB 2002

JAMB 2002 · UME · Q47

A circle with a radius of 5 cm has its radius increasing at the rate of 0.2 cm s−10.2\text{ cm s}^{-1}. What will be the corresponding rate of increase in the area?

  1. The formula

    • A=πr2{A = \pi r^2}, so dAdr=2πr{\frac{dA}{dr} = 2\pi r}.
    • When r=5{r = 5}: dAdr=10π{\frac{dA}{dr} = 10\pi}.

    Think first. Area of a circle, and its derivative.

  2. Connect the rates

    dAdt=10π×0.2=2π cm2s−1{\frac{dA}{dt} = 10\pi \times 0.2 = 2\pi\text{ cm}^2\text{s}^{-1}}: option C.

    Think first. Multiply by dr/dt = 0.2.

Velocity and acceleration

If ss is the distance travelled after time tt, the velocity is v=dsdtv = \dfrac{ds}{dt} and the acceleration is a=dvdta = \dfrac{dv}{dt}.

distance sd/dt →velocity vd/dt →acceleration a
differentiate to go right; integrate to come back
MotionDifferentiate distance to get velocity, and velocity to get acceleration

Your turn

JAMB 2018 · UTME · Q28

Find the value of xx for which the function f(x)=2x3−x2−4x+4f(x) = 2x^3 - x^2 - 4x + 4 has a maximum value.

Worked solution (try it first)
  1. At a turning point f′(x)=6x2−2x−4=0f'(x) = 6x^2 - 2x - 4 = 0.
  2. Factorise: 2(3x+2)(x−1)=02(3x + 2)(x - 1) = 0, so x=−23x = -\frac23 or x=1x = 1.
  3. f′′(x)=12x−2f''(x) = 12x - 2.
  4. At x=−23x = -\frac23 it is −10<0-10 < 0 (maximum).
  5. At x=1x = 1 it is 10>010 > 0 (minimum).
  6. So the maximum is at x=−23x = -\frac23, option C.

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