Piecewise models and continuity Problem set B Worked solutions

Find a so that \(f\) is continuous at x = 3: \(f(x)=\begin{cases}2x+a & x<3\\ x^2-4 & x\ge3\end{cases}\).

• Left at 3: \(2(3)+a=6+a\)
• Right at 3: \(3^2-4=5\)
• \(6+a=5\), so a = -1

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

• Left at 2: \(a(2)^2+2=4a+2\)
• Right at 2: \(3(2)-1=5\)
• \(4a+2=5\), so \(a=\tfrac{3}{4}\) = 0.75

Find a so that \(h\) is continuous at x = 4: \(h(x)=\begin{cases}x^a & 1\le x<4\\ 3x+6 & x\ge4\end{cases}\).

• Left at 4: \(4^a\). Right at 4: \(3(4)+6=18\)
• \(4^a=18\), so \(a\ln4=\ln18\)
• \(a=\ln18\div\ln4=2.08\)

Find a so that \(f\) is continuous at x = 3: \(f(x)=\begin{cases}(x+1)^a & x<3\\ 3x^2+2 & x\ge3\end{cases}\).

• Left at 3: \((3+1)^a=4^a\). Right at 3: \(3(3)^2+2=29\)
• \(4^a=29\), so \(a=\ln29\div\ln4=2.43\)

Find a so that \(h\) is continuous at x = 3: \(h(x)=\begin{cases}e^{ax} & x<3\\ x^2+5 & x\ge3\end{cases}\).

• Left at 3: \(e^{3a}\). Right at 3: \(3^2+5=14\)
• \(e^{3a}=14\), so \(3a=\ln14\)
• \(a=\ln14\div3=0.880\)

Find a so that \(g\) is continuous at x = 3: \(g(x)=\begin{cases}\ln(ax) & x<3\\ 2x-4 & x\ge3\end{cases}\).

• Left at 3: \(\ln(3a)\). Right at 3: \(2(3)-4=2\)
• \(\ln(3a)=2\), so \(3a=e^{2}\)
• \(a=e^{2}\div3=2.46\)

A tank holds water with 500 mg/l of a pollutant. Each day 9% of the pollutant is removed, so after n days the level is \(500\times0.91^n\) mg/l. Find the least whole number of days after which the level is less than 40 mg/l.

• \(500\times0.91^n<40\)
• Solve \(500\times0.91^n=40\): n = 26.8. Dividing by ln 0.91 (negative) turns < into >: n > 26.8
• Round up: n = 27 days

A phone plan costs, in pounds, \(C=12+0.9x\) for \(0\le x\le20\) GB, and \(C=ax+6\) for \(x>20\). The cost has no sudden jump at 20 GB. Find a.

• At 20 the first rule gives 12 + 0.9(20) = 30
• The second rule gives a(20) + 6
• 20 a + 6 = 30, so a = 1.2

The number of bacteria in a tank, in thousands, is \(900e^{-0.18d}\) after d days of treatment. Find the least whole number of days after which fewer than 50 thousand remain.

• \(900 e^{-0.18d}<50\)
• Solve \(900e^{-0.18d}=50\): d = 16.1
• The amount is below 50 from then on. Round up: 17 days

A car park charges £P for a stay of h hours as follows. Nothing for the first 2 hours. £2.50 for each hour after that, up to 5 hours in all. £4 for each hour after 5 hours.

1. (a) Find the charge for a stay of 7.5 hours.

   • The first 5 hours cost 2.5 × 3 = 7.5
   • The extra 2.5 hours cost 4 × 2.5 = 10
   • Total: £17.50

2. (b) Find the length of stay that costs £27.50.

   • The charge at 5 hours is £7.50. £27.50 is more, so use the last branch: 7.5 + 4(h − 5) = 27.5
   • Solve for h: h = 10

The function \(f\) is continuous at x = 3: \(f(x)=\begin{cases}e^{ax} & x<3\\ \ln(ax)+4 & x\ge3\end{cases}\). Find the two values of a.

• At x = 3: \(e^{3a}=\ln(3a)+4\)
• This cannot be rearranged. Solve on the calculator; graph \(y=e^{3x}-\ln(3x)-4\) to find both roots
• a = 0.0175 or a = 0.493
• Check a = 0.0175: \(e^{0.0525}=1.05\) and \(\ln(0.0525)+4=1.05\)

A patient takes 300 mg of a drug. The amount in the blood has a half-life of 5 hours. Find the least whole number of hours after which less than 2 mg remains.

• \(k=\ln2\div5=0.139\)
• \(300e^{-0.139t}<2\): solve \(300e^{-0.139t}=2\), t = 36.1
• Round up: 37 hours