Inverses and Radical Functions

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Show that \(f(x)=\frac{1}{x+1}\) and \({f}^{-1}(x)=\frac{1}{x}-1\) are inverses, for \(x\ne 0,-1\).

   Paljasta vastaus

   We must show that \({f}^{-1}(f(x))=x\) and \(f({f}^{-1}(x))=x.\)

   \[\begin{array}{l}{f}^{-1}(f(x))={f}^{-1}(\frac{1}{x+1}) \\ =\frac{1}{\frac{1}{x+1}}-1 \\ =(x+1)-1 \\ =x \\ f({f}^{-1}(x))=f(\frac{1}{x}-1) \\ =\frac{1}{(\frac{1}{x}-1)+1} \\ =\frac{1}{\frac{1}{x}} \\ =x\end{array}\]

   Therefore, \(f(x)=\frac{1}{x+1}\) and \({f}^{-1}(x)=\frac{1}{x}-1\) are inverses.

2. Show that \(f(x)=\frac{x+5}{3}\) and \({f}^{-1}(x)=3x-5\) are inverses.

   Paljasta vastaus

   \({f}^{-1}(f(x))={f}^{-1}(\frac{x+5}{3})=3(\frac{x+5}{3})-5=(x-5)+5=x\) and \(f({f}^{-1}(x))=f(3x-5)=\frac{(3x-5)+5}{3}=\frac{3x}{3}=x\)

3. Find the inverse of the function \(f(x)=5{x}^{3}+1.\)

   Paljasta vastaus

   This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for \(x.\)

   \[\begin{array}{l}\begin{array}{l} \\ y=5{x}^{3}+1\end{array} \\ x=5{y}^{3}+1 \\ x-1=5{y}^{3} \\ \frac{x-1}{5}={y}^{3} \\ {f}^{-1}(x)=\sqrt[3]{\frac{x-1}{5}}\end{array}\]

4. Find the inverse function of \(f(x)=\sqrt[3]{x+4}.\)

   Paljasta vastaus

   \({f}^{-1}(x)={x}^{3}-4\)

5. Find the inverse function of \(f:\)

   1. \(f(x)={(x-4)}^{2},\ x\ge 4\)
   2. \(f(x)={(x-4)}^{2},\ x\le 4\)
   Paljasta vastaus

   The original function \(f(x)={(x-4)}^{2}\) is not one-to-one, but the function is restricted to a domain of \(x\ge 4\) or \(x\le 4\) on which it is one-to-one. See .

   To find the inverse, start by replacing \(f(x)\) with the simple variable \(y.\)

   \[\begin{array}{ll}\ y={(x-4)}^{2} & \text{Interchange}\ x\ \text{and}\ y. \\ x={(y-4)}^{2} & \text{Take the square root}. \\ \pm \sqrt{x}=y-4\ & \text{Add}\ 4\ \text{to both sides}. \\ 4\pm \sqrt{x}=y & \end{array}\]

   This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of \(x\) and \(y\) for the original \(f(x),\) we looked at the domain: the values \(x\) could assume. When we reversed the roles of \(x\) and \(y,\) this gave us the values \(y\) could assume. For this function, \(x\ge 4,\) so for the inverse, we should have \(y\ge 4,\) which is what our inverse function gives.

   1. ⓐThe domain of the original function was restricted to \(x\ge 4,\) so the outputs of the inverse need to be the same, \(f(x)\ge 4,\) and we must use the + case: \[{f}^{-1}(x)=4+\sqrt{x}\]
   2. ⓑ The domain of the original function was restricted to \(x\le 4,\) so the outputs of the inverse need to be the same, \(f(x)\le 4,\) and we must use the – case: \[{f}^{-1}(x)=4-\sqrt{x}\]

6. Restrict the domain and then find the inverse of

   \[f(x)={(x-2)}^{2}-3.\]
   Paljasta vastaus

   We can see this is a parabola with vertex at \((2,\ -3)\) that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to \(x\ge 2.\)

   To find the inverse, we will use the vertex form of the quadratic. We start by replacing \(f(x)\) with a simple variable, \(y,\) then solve for \(x.\)

   \[\begin{array}{ll}\begin{array}{l} \\ y={(x-2)}^{2}-3\end{array} & \begin{array}{l} \\ \text{Interchange}\ x\ \text{and}\ y.\end{array} \\ x={(y-2)}^{2}-3 & \text{Add 3 to both sides}. \\ x+3={(y-2)}^{2} & \text{Take the square root}. \\ \pm \sqrt{x+3}=y-2 & \text{Add 2 to both sides}. \\ 2\pm \sqrt{x+3}=y & \text{Rename the function}. \\ {f}^{-1}(x)=2\pm \sqrt{x+3} & \end{array}\]

   Now we need to determine which case to use. Because we restricted our original function to a domain of \(x\ge 2,\) the outputs of the inverse should be the same, telling us to utilize the + case

   \[{f}^{-1}(x)=2+\sqrt{x+3}\]

   If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain.

7. Find the inverse of the function \(f(x)={x}^{2}+1,\) on the domain \(x\ge 0.\)

   Paljasta vastaus

   \({f}^{-1}(x)=\sqrt{x-1}\)

8. Restrict the domain and then find the inverse of the function \(f(x)=\sqrt{x-4}.\)

   Paljasta vastaus

   Note that the original function has range \(f(x)\ge 0.\) Replace \(f(x)\) with \(y,\) then solve for \(x.\)

   \[\begin{array}{lll}y & =\sqrt{x-4} & \text{Replace }f(x)\text{ with }y. \\ x & =\sqrt{y-4} & \text{Interchange }x\text{ and }y. \\ x & =\sqrt{y-4} & \text{Square each side}. \\ {x}^{2} & =y-4 & \text{Add 4}. \\ {x}^{2}+4 & =y & \text{Rename the function }{f}^{-1}(x). \\ {f}^{-1}(x) & ={x}^{2}+4 & \end{array}\]

   Recall that the domain of this function must be limited to the range of the original function.

   \[{f}^{-1}(x)={x}^{2}+4,x\ge 0\]

9. Restrict the domain and then find the inverse of the function \(f(x)=\sqrt{2x+3}.\)

   Paljasta vastaus

   \({f}^{-1}(x)=\frac{{x}^{2}-3}{2},x\ge 0\)

10. Park rangers construct a mound of gravel in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by

    \[V=\frac{2}{3}\pi {r}^{3}\]

    Find the inverse of the function \(V=\frac{2}{3}\pi {r}^{3}\) that determines the volume \(V\) of a cone and is a function of the radius \(r.\) Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use \(\pi =3.14.\)

    Paljasta vastaus

    Start with the given function for \(V.\) Notice that the meaningful domain for the function is \(r\ge 0\) since negative radii would not make sense in this context. Also note the range of the function (hence, the domain of the inverse function) is \(V\ge 0.\) Solve for \(r\) in terms of \(V,\) using the method outlined previously.
    

    \[\begin{array}{ll} & \\ V=\frac{2}{3}\pi {r}^{3} & \\ {r}^{3}=\frac{3V}{2\pi } & \text{Solve for }{r}^{3}. \\ r=\sqrt[3]{\frac{3V}{2\pi }} & \text{Solve for }r.\end{array}\]

    This is the result stated in the section opener. Now evaluate this for \(V=100\) and \(\pi =3.14.\)

    \[\begin{array}{l}\begin{array}{l} \\ r=\sqrt[3]{\frac{3V}{2\pi }}\ \end{array} \\ =\sqrt[3]{\frac{3⋅100}{2⋅3.14}} \\ \approx \sqrt[3]{47.7707} \\ \approx 3.63\end{array}\]

    Therefore, the radius is about 3.63 ft.

11. Find the domain of the function \(f(x)=\sqrt{\frac{(x+2)(x-3)}{(x-1)}}.\)

    Paljasta vastaus

    Because a square root is only defined when the quantity under the radical is non-negative, we need to determine where \(\frac{(x+2)(x-3)}{(x-1)}\ge 0.\) The output of a rational function can change signs (change from positive to negative or vice versa) at x-intercepts and at vertical asymptotes. For this equation, the graph could change signs at \(x\) = –2, 1, and 3.

    To determine the intervals on which the rational expression is positive, we could test some values in the expression or sketch a graph. While both approaches work equally well, for this example we will use a graph as shown in .

    This function has two x-intercepts, both of which exhibit linear behavior near the x-intercepts. There is one vertical asymptote, corresponding to a linear factor; this behavior is similar to the basic reciprocal toolkit function, and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator. There is a y-intercept at \((0,\sqrt{6}).\)

    From the y-intercept and x-intercept at \(x=-2,\) we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph.

    From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure that the original function \(f(x)\) will be defined. \(f(x)\) has domain \(-2\le x<1\ \text{or}\ x\ge 3,\) or in interval notation, \([-2,1)\cup [3,\infty ).\)

12. The function \(C=\frac{20+0.4n}{100+n}\) represents the concentration \(C\) of an acid solution after \(n\) mL of 40% solution has been added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for \(n\) in terms of \(C.\) Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution.

    Paljasta vastaus

    We first want the inverse of the function. We will solve for \(n\) in terms of \(C.\)

    \[\begin{array}{l}\ C=\frac{20+0.4n}{100+n} \\ C(100+n)=20+0.4n \\ 100C+Cn=20+0.4n \\ 100C-20=0.4n-Cn \\ 100C-20=(0.4-C)n \\ n=\frac{100C-20}{0.4-C}\end{array}\]

    Now evaluate this function for \(C=0.35\ (35\%).\)

    \[\begin{array}{l}\ n=\frac{100(0.35)-20}{0.4-0.35} \\ =\frac{15}{0.05} \\ =300\end{array}\]

    We can conclude that 300 mL of the 40% solution should be added.

13. Find the inverse of the function \(f(x)=\frac{x+3}{x-2}.\)

    Paljasta vastaus

    \({f}^{-1}(x)=\frac{2x+3}{x-1}\)

14. Explain why we cannot find inverse functions for all polynomial functions.

    Paljasta vastaus

    It can be too difficult or impossible to solve for \(x\) in terms of \(y.\)

15. Why must we restrict the domain of a quadratic function when finding its inverse?

16. When finding the inverse of a radical function, what restriction will we need to make?

    Paljasta vastaus

    We will need a restriction on the domain of the answer.

17. The inverse of a quadratic function will always take what form?

18. \(f(x)={(x-4)}^{2},\ [4,\infty )\)

    Paljasta vastaus

    \({f}^{-1}(x)=\sqrt{x}+4\)

19. \(f(x)={(x+2)}^{2},\ [-2,\infty )\)

20. \(f(x)={(x+1)}^{2}-3,\ [-1,\infty )\)

    Paljasta vastaus

    \({f}^{-1}(x)=\sqrt{x+3}-1\)

21. \(f(x)=2-\sqrt{3+x}\)

22. \(f(x)=3{x}^{2}+5,\ (-\infty ,0]\)

    Paljasta vastaus

    \({f}^{-1}(x)=-\sqrt{\frac{x-5}{3}}\)

23. \(f(x)=12-{x}^{2},\ [0,\infty )\)

24. \(f(x)=9-{x}^{2},\ [0,\infty )\)

    Paljasta vastaus

    \(f(x)=\sqrt{9-x}\)

25. \(f(x)=2{x}^{2}+4,\ [0,\infty )\)

26. \(f(x)={x}^{3}+5\)

    Paljasta vastaus

    \({f}^{-1}(x)=\sqrt[3]{x-5}\)

27. \(f(x)=3{x}^{3}+1\)

28. \(f(x)=4-{x}^{3}\)

    Paljasta vastaus

    \({f}^{-1}(x)=\sqrt[3]{4-x}\)

29. \(f(x)=4-2{x}^{3}\)

30. \(f(x)=\sqrt{2x+1}\)

    Paljasta vastaus

    \({f}^{-1}(x)=\frac{{x}^{2}-1}{2},\ [0,\infty )\)

31. \(f(x)=\sqrt{3-4x}\)

32. \(f(x)=9+\sqrt{4x-4}\)

    Paljasta vastaus

    \({f}^{-1}(x)=\frac{{(x-9)}^{2}+4}{4},\ [9,\infty )\)

33. \(f(x)=\sqrt{6x-8}+5\)

34. \(f(x)=9+2\sqrt[3]{x}\)

    Paljasta vastaus

    \({f}^{-1}(x)={(\frac{x-9}{2})}^{3}\)

35. \(f(x)=3-\sqrt[3]{x}\)

36. \(f(x)=\frac{2}{x+8}\)

    Paljasta vastaus

    \({f}^{-1}(x)={\frac{2-8x}{x}}^{}\)

37. \(f(x)=\frac{3}{x-4}\)

38. \(f(x)=\frac{x+3}{x+7}\)

    Paljasta vastaus

    \({f}^{-1}(x)=\frac{7x-3}{1-x}\)

39. \(f(x)=\frac{x-2}{x+7}\)

40. \(f(x)=\frac{3x+4}{5-4x}\)

    Paljasta vastaus

    \({f}^{-1}(x)=\frac{5x-4}{4x+3}\)

Kokeile omaasi

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Lisää Precalculus

Complex numbersPolynomial functionsRational functionsSequences and seriesThe binomial theoremConic sectionsVectorsExponential and logarithmic functionsPolynomial division and the remainder theoremParametric equations and polar coordinates