Inverse functions Problem set B Worked solutions

Find the inverse function of each function.

1. (a) \(f(x)=5 x + 9\)

   • \(y=5 x + 9\)
   • \(x=\frac{y - 9}{5}\)
   • \(f^{-1}(x)=\frac{x - 9}{5}\)

2. (b) \(g(x)=\frac{7 - 4 x}{3}\)

   • \(y=\frac{7 - 4 x}{3}\)
   • \(x=\frac{7 - 3 y}{4}\)
   • \(g^{-1}(x)=\frac{7 - 3 x}{4}\)

\(f(x)=3 x - 11\) and \(g(x)=\frac{x + 11}{3}\). Show that g is the inverse of f.

1. (a) \(f(g(x))=x\)

   • \(f(g(x))=3 \left(\frac{x + 11}{3}\right) - 11\)
   • \(f(g(x))=x\)

2. (b) \(g(f(x))=x\)

   • \(g(f(x))=\frac{\left(3 x - 11\right) + 11}{3}\)
   • \(g(f(x))=x\)

The function \(f(x)=x^{3} - 9 x^{2} + 4\) is not one-to-one for all x. Use your calculator to find the largest domain of the form \(x \geq k\) for which f has an inverse.

• Graph f and look for the turning point.
• From the calculator: minimum at (6, −104)
• To the right of the minimum, f is increasing: one-to-one.
• \(x \geq 6\)

The graph shows \(y=e^x\), \(y=\ln x\) and the line \(y=x\). The point \(B(3, 1.10)\) lies on \(y=\ln x\).

1. (a) Write down the coordinates of the point on \(y=e^x\) that is the reflection of B.

   • The two graphs are reflections of each other in y = x.
   • Swap the coordinates: (1.10, 3)

2. (b) State the equation of the mirror line.

   • y = x

The graph of \(y=f(x)\) passes through the points \((0, 1)\), \((2, 7)\), \((3, 10)\) and \((5, 16)\). The function has an inverse. Write down the coordinates of the matching points on the graph of \(y=f^{-1}(x)\).

• The graph of \(f^{-1}\) is the reflection of the graph of \(f\) in the line \(y=x\).
• A reflection in \(y=x\) swaps the coordinates: \((a, b)\) becomes \((b, a)\).
• (1, 0), (7, 2), (10, 3), (16, 5)

The graph shows \(y=f(x)\). The point \((2, 9)\) lies on the graph.

1. (a) Write down \(f(2)\).

   • Read up from \(x=2\) to the curve: \(y=9\).

2. (b) Write down \(f^{-1}(27)\).

   • \(f^{-1}(27)\) is the x-value where \(f(x)=27\).
   • Read across from \(y=27\) to the curve: \(x=3\).

3. (c) Write down the point on the graph of \(y=f^{-1}(x)\) that matches \((2, 9)\).

   • Swap the coordinates of \((2, 9)\).
   • (9, 2)

The graph shows \(y=10^x\), \(y=\log_{10}x\) and the line \(y=x\). The point \(Q(5, 0.699)\) lies on \(y=\log_{10}x\).

1. (a) Write down the coordinates of the point on \(y=10^x\) that is the reflection of Q.

   • \(10^x\) and \(\log_{10}x\) are inverses, so their graphs are reflections in \(y=x\).
   • Swap the coordinates: (0.699, 5)

2. (b) Use your calculator to find \(10^{0.699}\). What does your answer tell you about the point \((0.699, 5)\)?

   • From the calculator: \(10^{0.699}=5.00\)
   • So \((0.699, 5)\) lies on \(y=10^x\).

Find the inverse of each function. State the domain of the inverse.

1. (a) \(f(x)=4e^x-3\)

   • \(y=4e^x-3\)
   • \(e^x=\frac{y+3}{4}\), so \(x=\ln\!\left(\frac{y+3}{4}\right)\)
   • The range of f is \(y>-3\), so the domain of \(f^{-1}\) is \(x>-3\).

2. (b) \(g(x)=2\ln x+5\), \(x>0\)

   • \(y=2\ln x+5\)
   • \(\ln x=\frac{y-5}{2}\), so \(x=e^{(y-5)/2}\)
   • The range of g is all real numbers, so the domain of \(g^{-1}\) is all real x.

The function f is defined by \(f(x)=x^2+10x+31\), for \(x\geq-5\).

1. (a) Write f in the form \((x-h)^2+k\).

   • \(x^2+10x+31=(x+5)^2-25+31\)
   • \(f(x)=(x+5)^2+6\)

2. (b) Find \(f^{-1}(x)\).

   • \(y=(x+5)^2+6\)
   • \(x+5=\sqrt{y-6}\) (positive root, because \(x\geq-5\))
   • \(f^{-1}(x)=-5+\sqrt{x-6}\)

3. (c) State the domain and the range of \(f^{-1}\).

   • The range of f is \(y\geq6\), so the domain of \(f^{-1}\) is \(x\geq6\).
   • The domain of f is \(x\geq-5\), so the range of \(f^{-1}\) is \(y\geq-5\).

A taxi fare is c(d) = 2.5 + 1.5d pounds for a journey of d kilometres.

1. (a) Find the inverse function, and say what it tells you.

   • \(y=\frac{3 x + 5}{2}\)
   • \(x=\frac{2 y - 5}{3}\)
   • \(c^{-1}(x)=\frac{2 x - 5}{3}\)

2. (b) Use it to find the distance for a fare of £20, in kilometres to 3 significant figures.

   • \(c^{-1}(20)=11.7\)

A tank holds V = f(t) = 40 + 2.5t litres of water after t minutes. Mia says the inverse is f⁻¹(x) = 0.4x − 40.

1. (a) Work out \(f(f^{-1}(x))\) using this inverse, and say whether it is correct.

   • \(f(f^{-1}(x))=f(0.4 x - 40)\)
   • \(=x - 60\)
   • A correct inverse gives x. This is not the inverse.

2. (b) Write down the correct inverse.

   • \(y=2.5 x + 40\)
   • \(x=0.4 y - 16\)
   • \(f^{-1}(x)=0.4 x - 16\)

A taxi company charges by the distance travelled. The fare \(F(d)\) pounds for a journey of \(d\) kilometres is shown in the table.

1. (a) Write down \(F(6)\).

   • Read the table at \(d=6\): \(F(6)=15\).

2. (b) Find \(F^{-1}(19)\) and say what it means.

   • \(F^{-1}(19)\) is the distance that costs a fare of £19.
   • In the table, \(F(8)=19\), so \(F^{-1}(19)=8\).
   • A fare of £19 is for a journey of 8 km.

3. (c) Write down the point on the graph of \(y=F^{-1}(x)\) that matches \((10, 23)\).

   • Swap the coordinates of \((10, 23)\).
   • (23, 10)

The graph shows a straight line \(y=g(x)\) that converts an amount of \(x\) pounds into \(y\) dollars. It passes through \((0, 0)\) and \((40, 52)\).

1. (a) Write down the coordinates of the point on the graph of \(y=g^{-1}(x)\) that matches \((40, 52)\).

   • The graph of \(g^{-1}\) is the reflection of the graph of \(g\) in \(y=x\).
   • Swap the coordinates of \((40, 52)\): (52, 40)

2. (b) What does this point tell you?

   • \(g(40)=52\), so \(g^{-1}(52)=40\).
   • 52 dollars is £40. \(g^{-1}\) turns dollars into pounds.

\(f(x)=\frac{6}{x + 2}\), \(x \neq -2\).

1. (a) Find \(f^{-1}(x)\).

   • \(y=\frac{6}{x + 2}\)
   • \(x=\frac{2 \left(3 - y\right)}{y}\)
   • \(f^{-1}(x)=\frac{2 \left(3 - x\right)}{x}\)

2. (b) Find \(f^{-1}(8)\).

   • \(f^{-1}(8)=-1.25\)

The function \(f(x)=x^{3} e^{- x}\), \(x \geq 0\), is not one-to-one. Use your calculator to find the largest domain of the form \(x \geq k\) for which f has an inverse.

• Graph f and look for the turning point.
• From the calculator: maximum at (3, 1.34)
• To the right of the maximum f is decreasing, so it is one-to-one.
• \(x \geq 3\)

The graph of \(y=f(x)\), where \(f(x)=0.4x+3\), is reflected in the line \(y=x\) to give the graph of \(y=f^{-1}(x)\).

1. (a) Use your calculator to find the point where \(y=f(x)\) meets \(y=x\).

   • Solve \(0.4x+3=x\): \(x=5\).
   • \(y=x=5\), so the point is (5, 5).

2. (b) Write down the point where \(y=f^{-1}(x)\) meets \(y=x\). Explain your answer.

   • Reflecting in \(y=x\) swaps the coordinates.
   • For (5, 5) they are the same, so the point does not move.
   • (5, 5)

The function f is defined by \(f(x)=3\ln x+0.5\), for \(x>0\).

1. (a) Find \(f^{-1}(x)\).

   • \(y=3\ln x+0.5\)
   • \(\ln x=\frac{y-0.5}{3}\)
   • \(f^{-1}(x)=e^{(x-0.5)/3}\)

2. (b) Solve \(f(x)=f^{-1}(x)\). Give each solution to 3 significant figures.

   • The graphs of f and \(f^{-1}\) meet on the line \(y=x\), so solve \(3\ln x+0.5=x\).
   • From the calculator: \(x=1.31\) and \(x=5.74\)