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AI HL · 12.7

Mathematical Models

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12.7.5

Piecewise models and continuity

12.7.5

Piecewise models and continuity

Target: I can find an unknown a so that a piecewise function is continuous, by putting both branches equal at the boundary. (+4 more)

By the end of this section

Key points

Continuous piecewise functions

  • A function is continuous at \(x=x_0\) if its graph has no break there: the two branches meet at the same point.
  • Put \(x=x_0\) into each branch. Set the two values equal and solve for the unknown.
  • Use the boundary value in both branches, even when the inequality is strict on one side.
  • Check: put a back in and confirm both branches give the same value at \(x_0\).

When a is in the exponent or a logarithm

  • If \(x^a=v\), take ln of both sides: \(a\ln x=\ln v\), so \(a=\dfrac{\ln v}{\ln x}\).
  • The same works for \((x+1)^a\): put the boundary in first, then take ln.
  • If \(e^{ax_0}=v\), then \(ax_0=\ln v\). If \(\ln(ax_0)=v\), then \(ax_0=e^{v}\).
  • If a cannot be rearranged out, solve the equation on the calculator.

Decay and whole days

  • Decay by a fixed percentage each day: after n days the amount is \(A_0\times r^n\), where r is the daily multiplier (6% loss: r = 0.94).
  • Write the inequality, such as \(80\times0.94^n<10\). Solve it as an equation first, with logs or the calculator.
  • Dividing by ln r, a negative number, reverses the inequality. Graph the model to check which side is right.
  • 'The least whole number of days' means round up to the next whole number, not to the nearest.

Stepped-rate models

  • A tariff or a tax charges one rate up to a limit and another rate on the part above it. Write each branch for its own interval of x.
  • Only the part above a limit is charged at the new rate. For \(x>200\) the charge is the cost of the first 200, plus the new rate × \((x-200)\).
  • To find x for a given output: work out the output at each limit first, and choose the branch the output falls in. Then solve in that branch.
  • Check the answer is inside the interval of the branch you used.

On the calculator

Solving for a on the calculator

  • Type the equation with a as the unknown, for example \(e^{2a}=\ln(2a)+3\).
  • Solve for a. If the equation may have more than one solution, use the graph of the difference to find them all.
  • Keep a in the calculator and check both branches give the same value at the boundary.

Screenshot and video to come

Solving a decay inequality

  • Graph the model and the line at the target value.
  • Find the intersection. The solution is every n beyond it.
  • Round up to the next whole number of days.

Screenshot and video to come

Explore: when do two branches meet?

The orange branch contains the unknown a. Work out the value of a that makes the model continuous, type it in, then check the graph. A filled circle means the point belongs to that branch. An open circle means it does not.



Value of a needed

Move the slider or type a value.

  1. Before you touch the slider, put the boundary value of x into both branches and work out a. Then type it in. Does the graph agree?
  2. Look at the circles at the boundary. Which branch includes the boundary point, and how can you tell? What happens to the circles when the model is continuous?
  3. On the (x + 1)a model you cannot find a by dividing. What do you do instead, and why does it work?
  4. Try a value that is close to the right one but not equal. Does the model count as continuous? What does the gold bar show?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Finding a for continuity

Find a so that \(f\) is continuous at x = 2: \(f(x)=\begin{cases}ax^2+3 & x\le2\\ 5x-1 & x>2\end{cases}\).

0.511.522.533.545101520xy
Working
  1. Left branch at 2

    \(a(2)^2+3=4a+3\)

    Put x = 2 in the first branch.

  2. Right branch at 2

    \(5(2)-1=9\)

    Put x = 2 in the second branch.

  3. Set equal

    \(4a+3=9\), so \(a=\tfrac32\)

    The two pieces must meet at the same height.

  4. Check

    Left: 4(1.5) + 3 = 9. Right: 9.

    Both give 9, so the graph has no break.

Video walkthrough coming
When a is the exponent

Find a so that \(g\) is continuous at x = 3: \(g(x)=\begin{cases}(x+1)^a & 0\le x\le3\\ 7x-1 & x>3\end{cases}\).

Working
  1. Both branches at 3

    \((3+1)^a=4^a\) and \(7(3)-1=20\)

    Put in the boundary first.

  2. Set equal

    \(4^a=20\)

    a is now the unknown power.

  3. Take ln

    \(a\ln4=\ln20\), so \(a=\dfrac{\ln20}{\ln4}=2.16\)

    Or solve on the calculator.

Video walkthrough coming
When a is inside e to the power

Find a so that \(h\) is continuous at x = 2: \(h(x)=\begin{cases}e^{ax} & x<2\\ 3x+1 & x\ge2\end{cases}\).

Working
  1. At x = 2

    \(e^{2a}\) and \(3(2)+1=7\)

    Put x = 2 in both.

  2. Set equal

    \(e^{2a}=7\)

    a is in the power.

  3. Take ln

    \(2a=\ln7\), so \(a=0.973\)

    ln undoes e.

Video walkthrough coming
When a is inside a logarithm

Find a so that \(g\) is continuous at x = 5: \(g(x)=\begin{cases}\ln(ax) & x<5\\ 2x-7 & x\ge5\end{cases}\).

Working
  1. At x = 5

    \(\ln(5a)\) and \(2(5)-7=3\)

    Put x = 5 in both.

  2. Set equal

    \(\ln(5a)=3\)

    a is inside the ln.

  3. Undo the ln

    \(5a=e^{3}\), so \(a=\dfrac{e^3}{5}=4.02\)

    e to the power of each side.

Video walkthrough coming
A decay inequality

A pond holds 500 mg of a chemical. Each day 8% of the chemical breaks down, so after n days the amount is \(500\times0.92^n\) mg. Find the least whole number of days after which less than 40 mg remains.

Working
  1. The inequality

    \(500\times0.92^n<40\)

    Less than 40 mg.

  2. Solve as an equation

    Solve for n: \(500\times0.92^n=40\): n = 30.3

    Or: \(n=\ln0.08\div\ln0.92\). Dividing by ln 0.92, which is negative, turns < into >: n > 30.3.

  3. Whole days

    n = 31

    The amount is below 40 mg from day 31 onwards. Check: 500 × 0.92^31 = 37.7.

Video walkthrough coming
An electricity bill

An electricity company charges £0.18 for each kWh for the first 200 kWh in a month, and £0.25 for each kWh after that. The bill is £B for x kWh. (a) Write B as a piecewise function of x. (b) Find the bill for 350 kWh. (c) Find the number of kWh used when the bill is £87.

Working
  1. The first branch

    \(B=0.18x,\ 0\le x\le200\)

    The first 200 kWh.

  2. The second branch

    \(B=0.18\times200+0.25(x-200)=36+0.25(x-200),\ x>200\)

    The first 200 kWh cost £36. Add the new rate on the rest.

  3. Part (b)

    \(B=36+0.25\times150=73.5\)

    350 − 200 = 150 kWh at the higher rate: £73.50.

  4. Part (c)

    \(36+0.25(x-200)=87\)

    £87 is more than £36, so use the second branch. Solve for x: x = 404. This is more than 200, so the branch is right.

Video walkthrough coming