Exercise (4.7) - Inverse Functions

  1. Find the formula for $f^{-1}$ and state the domain of $f^{-1}$ when the function $f$ is given by

    (a) $f(x)=2 x-3$


    $\begin{array}{l} f(x)=2x-3\\\\ \text{Let}\ f^{-1}(x)=y\ \text{then}\ f(y)=x\\\\ \therefore\quad 2y - 3 = x\\\\ \quad\quad y=\displaystyle\frac{x+3}{2}\\\\ \therefore\quad f^{-1}(x)=\displaystyle\frac{x+3}{2} \end{array}$

    (b) $f(x)=1+3 x$


    $\begin{array}{l} f(x)=1+3x\\\\ \text{Let}\ f^{-1}(x)=y\ \text{then}\ f(y)=x\\\\ \therefore\quad 1+3y = x\\\\ \quad\quad y=\displaystyle\frac{x-1}{3}\\\\ \therefore\quad f^{-1}(x)=\displaystyle\frac{x-1}{3} \end{array}$

    (c) $f(x)=1-x$


    $\begin{array}{l} f(x)=1-x\\\\ \text{Let}\ f^{-1}(x)=y\ \text{then}\ f(y)=x\\\\ \therefore\quad 1-y = x\\\\ \quad\quad y=1-x\\\\ \therefore\quad f^{-1}(x)=1-x \end{array}$

    (d) $f(x)=\displaystyle\frac{x+9}{2}$


    $\begin{array}{l} f(x)=\displaystyle\frac{x+9}{2}\\\\ \text{Let}\ f^{-1}(x)=y\ \text{then}\ f(y)=x\\\\ \therefore\quad \displaystyle\frac{y+9}{2} = x\\\\ \quad\quad y=2x-9\\\\ \therefore\quad f^{-1}(x)=2x-9 \end{array}$

    (e) $f(x)=\displaystyle\frac{1}{3}(4 x-5)$


    $\begin{array}{l} f(x)=\displaystyle\frac{1}{3}(4x-5)\\\\ \text{Let}\ f^{-1}(x)=y\ \text{then}\ f(y)=x\\\\ \therefore\quad \displaystyle\frac{1}{3}(4y-5) = x\\\\ \quad\quad y=\displaystyle\frac{3x+5}{4}\\\\ \therefore\quad f^{-1}(x)=\displaystyle\frac{3x+5}{4} \end{array}$

    (f) $f(x)=\displaystyle\frac{2 x+5}{x-7}$


    $\begin{array}{l} f(x)=\displaystyle\frac{2x+5}{x-7}\\\\ \text{Let}\ f^{-1}(x)=y\ \text{then}\ f(y)=x\\\\ \therefore\quad \displaystyle\frac{2y+5}{y-7}= x\\\\ \quad\quad 2y+5 = xy-7x\\\\ \quad\quad xy-2y= 7x-5\\\\ \quad\quad y(x-2)= 7x-5\\\\ \quad\quad y=\displaystyle\frac{7x-5}{x-2}\\\\ \therefore\quad f^{-1}(x)=\displaystyle\frac{7x-5}{x-2} \end{array}$

    (g) $f(x)=\displaystyle\frac{3}{x-2}$


    $\begin{array}{l} f(x)=\displaystyle\frac{3}{x-2}\\\\ \text{Let}\ f^{-1}(x)=y\ \text{then}\ f(y)=x\\\\ \therefore\quad \displaystyle\frac{3}{y-2}= x\\\\ \quad\quad y-2 = \displaystyle\frac{3}{x}\\\\ \quad\quad y= \displaystyle\frac{3}{x}+2\\\\ \quad\quad y=\displaystyle\frac{2x+3}{x}\\\\ \therefore\quad f^{-1}(x)=\displaystyle\frac{2x+3}{x} \end{array}$

    (h) $f(x)=\displaystyle\frac{13}{2 x}$


  2. $\begin{array}{l} f(x)=\displaystyle\frac{13}{2x}\\\\ \text{Let}\ f^{-1}(x)=y\ \text{then}\ f(y)=x\\\\ \therefore\quad \displaystyle\frac{13}{2y}= x\\\\ \quad\quad y = \displaystyle\frac{13}{2x}\\\\ \therefore\quad f^{-1}(x)=\displaystyle\frac{13}{2x} \end{array}$

  3. $A=\{x \mid x \geq 0, x \in \mathbb{R}\}$ and $g, h$ are functions from $A$ to $A$ defined by $g(x)=$ $2 x, h(x)=x^{2}$

    (a) Find the formula for the inverse functions $g^{-1}, h^{-1}$.

    (b) Evaluate $g^{-1}(7), h^{-1}(5)$.


  4. $\begin{array}{l} A=\{x \mid x \geq 0, x \in \mathbb{R}\}\\\\ g(x)=2x\\\\ h(x)=x^2\\\\ \text{Let}\ g^{-1}(x)=y\ \text{then}\ g(y)=x\\\\ \therefore\quad 2y= x\\\\ \quad\quad y=\displaystyle\frac{x}{2}\\\\ \therefore\quad g^{-1}(x)=\displaystyle\frac{x}{2}\\\\ \text{Let}\ h^{-1}(x)=z\ \text{then}\ h(z)=x\\\\ \therefore\quad z^2= x\\\\ \quad\quad z=\sqrt{x}\\\\ \therefore\quad h^{-1}(x)=\sqrt{x}\\\\ \quad\quad g^{-1}(7)=\displaystyle\frac{7}{2}\\\\ \quad\quad h^{-1}(9)=\sqrt{9}=3 \end{array}$

  5. Function $f$ is given by $f(x)=\displaystyle\frac{2 x-5}{x-3}$.

    (a) State the value of $x$ for which $f$ is not defined.

    (b) Find the value of $x$ for which $f(x)=0$.

    (c) Find the inverse function $f^{-1}$ and state the domain of $f^{-1}$.


  6. $\begin{array}{ll} \text{(a)} & f(x)=\displaystyle\frac{2x-5}{x-3}\\\\ & f(x)\ \text{is not defined when}\\\\ & x-3 = 0\ \text{or}\ x=3 \\\\ \text{(b)} & f(x)=0\\\\ & \displaystyle\frac{2x-5}{x-3}=0\\\\ & \therefore\quad 2x-5=0\\\\ & \quad\quad x=\frac{5}{2}\\\\ \text{(c)} & \text{Let}\ f^{-1}(x)=y\ \text{then}\ f(y)=x\\\\ & \therefore\quad \displaystyle\frac{2y-5}{y-3}= x\\\\ & \quad\quad 2y-5=xy-3x\\\\ & \quad\quad xy-2y=3x-5\\\\ & \quad\quad y(x-2)=3x-5\\\\ & \quad\quad y=\displaystyle\frac{3x-5}{x-2}, x\ne 2\\\\ & \therefore\quad f^{-1}=\displaystyle\frac{3x-5}{x-2}, x\ne 2\\\\ & \therefore\quad \text{Domain of}\ f^{-1}=\left\{x|\ x\ne 2, x \in \mathbb{R} \right\} \end{array}$

  7. Function $f$ is given by $f(x)=\displaystyle\frac{x+a}{x-2}$ and that $f(7)=2$, find

    (a) the value of $a$, and

    (b) $f^{-1}(-4)$.


  8. $\begin{array}{ll} \text{(a)} & f(x)=\displaystyle\frac{x+a}{x-2}\\\\ & f(7)=2\\\\ & \displaystyle\frac{7+a}{7-2}=2\\\\ & a=3\\\\ \text{(b)} & \text{Let}\ f^{-1}(-4)=k\ \text{then}\ f(k)=-4\\\\ & \displaystyle\frac{k+3}{k-2}=-4\\\\ & k+3=-4k+8\\\\ & 5k=5\\\\ & \therefore\quad k=1 \end{array}$

  9. The function $f$ is given by $f(x)=4^{x}-2$.

    (a) Find the value of $x$ for which $f(x)=0$.

    (b) Find the inverse function $f^{-1}$ and state the domain of $f^{-1}$.

    (c) If $f^{-1}(k)=2$, find the value of $k$.


  10. $\begin{array}{ll} \text{(a)} & f(x)=4^x-2\\\\ & f(x)=0\\\\ & 4^x-2=0\\\\ & 4^x=2\\\\ & 4^x=4^{\frac{1}{2}}\\\\ & x=\displaystyle\frac{1}{2}\\\\ \text{(b)} & \text{Let}\ f^{-1}(x)=y\ \text{then}\ f(x)=y\\\\ & 4^y-2=x\\\\ & 4^y=x+2\\\\ & y=\log_{4}{x+2}\\\\ & \therefore\quad f^{-1}(x)=\log_{4}{x+2}\\\\ & f^{-1}(x)\ \text{is defined only when}\ x+2\ge 0,i.e., x\ge -2\\\\ & \therefore\quad \text{Domain of}\ f^{-1}=\left\{x|\ x\ge -2, x \in \mathbb{R} \right\}\\\\ \text{(c)} & f^{-1}(k)=2\\\\ & \therefore\quad k=f(2)\\\\ & k=4^2-2 = 14 \end{array}$