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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 are 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 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=0f'(x) = 3x^2 - 12 = 0, so x2=4x^2 = 4 and x=2x = 2 or x=−2x = -2.

    Think first. Solve f′(x) = 0.

  2. Which is the maximum?

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

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

  3. The value

    f(−2)=−8+24+11=27f(-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, write the quantity as a function of one variable, differentiate, set the derivative to 0 and solve.

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 the 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=πr2A = \pi r^2, so dAdr=2πr=10π\frac{dA}{dr} = 2\pi r = 10\pi when r=5r = 5.

    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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