Chapter 1.5, Problem 46E

Solution

(a)

To find: $f(x)$ has graph with horizontal asymptote but $f^{-1}(x)$ don’t.

  $f(x)=\frac{1}{x^2+2}$ has horizontal asymptote but its inverse $f^{-1}(x)=\sqrt{\frac{1}{x}-2}$ don’t have horizontal asymptote.

Given informationLet the function be $f(x)=\frac{1}{x^2+2}$ and its inverse is $f^{-1}(x)=\sqrt{\frac{1}{x}-2}$ .

The graph of both functions given below:

  

As It is observed from the graph $f(x)=\frac{1}{x^2+2}$ has horizontal asymptote as $x$ axis but its inverse don’t have the horizontal asymptote.

(b)

To find: $f(x)$ has domain of all real number but $f^{-1}(x)$ don’t.

The function $y=x^2$ has domain all real number but $y=\sqrt{x}$ restricted domain $\left[0,\infty \right)$ .

Given information Let the function be $y=x^2$ .The domain of the function is all real number.

The inverse of above function is $y=\sqrt{x}$ in restricted domain $\left[0,\infty \right)$ .the function is not defined for negative values of $x$ .Hence it can be said that $f(x)$ has domain of all real number but $f^{-1}(x)$ don’t.

(c)

To find: $f(x)$ has the graph that bounded above but $f^{-1}(x)$ don’t.

The function $y=2-x^2$ is bounded above but $y=\sqrt{2-x}$ don’t.

Given information Let the function be $y=2-x^2$ .The inverse of this function is $\sqrt{2-x}$

.The graph of both function is given below:

  

The function $y=2-x^2$ is bounded above and the inverse $\sqrt{2-x}$ is bounded below by the $x-axis$ .

(d)

To find: $f(x)$ has removable discontinuity at $x=5$ but $f^{-1}(x)$ don’t.

The function $y=\frac{x^2-25}{x-5}$ has removable discontinuity at $x=5$ but $f^{-1}(x)$ don’t.

Given information Let the function be $y=\frac{x^2-25}{x-5}$ has removable discontinuity at $x=5$ .The inverse of this function is $\frac{x^2-25}{x+5}$ . $f(x)$ has removable discontinuity at $x=5$ but $f^{-1}(x)$ has removable discontinuity at $x=-5$ .