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Inverses and Radical Functions

Find the inverse of a polynomial function.

Finding the Inverse of a Polynomial Function

Two functions \(f\) and \(g\) are inverse functions if for every coordinate pair in \(f,\ (a,\ b),\) there exists a corresponding coordinate pair in the inverse function, \(g,(b,\ a).\) In other words, the coordinate pairs of the inverse functions have the input and output interchanged.

For a function to have an inverse, it must be one-to-one.

For example, suppose the Sustainability Club builds a water runoff collector in the shape of a parabolic trough We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.

Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with \(x\) measured horizontally and \(y\) measured vertically, with the origin at the vertex of the parabola.

From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form \(y(x)=a{x}^{2}.\) Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor \(a.\)

\[\begin{array}{l} \\ 18=a{6}^{2} \\ a=\frac{18}{36} \\ =\frac{1}{2}\end{array}\]

Our parabolic cross section has the equation

\[y(x)=\frac{1}{2}{x}^{2}\]

We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth \(y\) the width will be given by \(2x,\) so we need to solve the equation above for \(x\) and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.

\[\begin{array}{l}\ y=\frac{1}{2}{x}^{2} \\ 2y={x}^{2} \\ x=\pm \sqrt{2y} \\ ∴y=\pm \sqrt{2x}\end{array}\]\[y=\sqrt{2x}\]\[\begin{array}{l}\text{Area}=l⋅w \\ =36⋅2x \\ =72x \\ =72\sqrt{2y}\end{array}\]
  1. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.
  2. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.
\[{f}^{-1}(f(x))=x,\ \text{for all }x\text{ in the domain of }f\]\[f({f}^{-1}(x))=x,\ \text{for all }x\text{ in the domain of }{f}^{-1}\]

Condensed: the full section is in OpenStax Precalculus 2e.

Restricting the Domain to Find the Inverse of a Polynomial Function

So far, we have been able to find the inverse functions of cubic functions without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an inverse function. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses.

Example

Try it.

Find the inverse function of \(f:\)

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

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.

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

Condensed: the full section is in OpenStax Precalculus 2e.

Key Concepts

  • The inverse of a quadratic function is a square root function.
  • If \({f}^{-1}\) is the inverse of a function \(f,\) then \(f\) is the inverse of the function \({f}^{-1}.\)
  • While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible.
  • To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one.
  • When finding the inverse of a radical function, we need a restriction on the domain of the answer.
  • Inverse and radical and functions can be used to solve application problems.

Inverses and Radical Functions

  • Given function, find the inverse function (IA 10.1.3)
  • Find the domain of a radical function (IA 8.7.2)

For example, \(f(x)=\sqrt{x+3}\) , \(g(x)=\sqrt[3]{2x-1}\) are both radical functions.

Try it.

For the radical function \(f(x)=\sqrt{x-2}\) , find the following

  1. ⓐ \(f(3)\)
  2. ⓑ \(f(6)\)
  3. ⓒ \(f(2)\)
  4. ⓓ \(f(1)\)
  5. ⓔ Why is there a problem with part d? Explain.
  6. ⓕ What do you think is the domain of \(f(x)\) ? (Hint: think of domain as all the x-values for which we can evaluate \(f(x)\)

Try it.

For the radical function \(f(x)=\sqrt[3]{x-2}\) , find the following

  1. ⓐ \(f(3)\)
  2. ⓑ \(f(6)\)
  3. ⓒ \(f(2)\)
  4. ⓓ \(f(1)\)
  5. ⓔ Why is there a problem with part d? Explain.
  6. ⓕ What do you think is the domain of \(f(x)\) ? (Hint: think of domain as all the x-values for which we can evaluate \(f(x)\)
Example

Find the domain of a radical function.

Try it.

Find the domain of the function \(f(x)=\sqrt{3x-4}\) . Write the domain in interval notation.

Solution

Since this function has a radical with the even index of 2, we set the radicand to be greater than or equal to 0 and then solve to find the domain.

\(\begin{array}{l}3x-4\ge 0 \\ 3x\ge 4 \\ x\ge \frac{4}{3}\end{array}\)

The domain expressed in interval notation is \([\frac{4}{3},\infty ]\)

Find the domain of the following functions and express using interval notation.

Try it.

\(f(x)=\sqrt{3x+5}\)

Try it.

\(f(x)=\sqrt[3]{4-6x}\)

Try it.

\(f(x)=\sqrt[4]{5-x}\)

Condensed: the full section is in OpenStax College Algebra 2e.

Finding the Inverse of a Polynomial Function

Two functions \(f\) and \(g\) are inverse functions if for every coordinate pair in \(f,(a,b),\) there exists a corresponding coordinate pair in the inverse function, \(g,(b,\ a).\) In other words, the coordinate pairs of the inverse functions have the input and output interchanged. Only one-to-one functions have inverses. Recall that a one-to-one function has a unique output value for each input value and passes the horizontal line test.

For example, suppose the Sustainability Club builds a water runoff collector in the shape of a parabolic trough We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.

Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with \(x\) measured horizontally and \(y\) measured vertically, with the origin at the vertex of the parabola.

From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form \(y(x)=a{x}^{2}.\) Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor \(a.\)

\[\begin{array}{lll}18 & = & a{6}^{2} \\ a & = & \frac{18}{36} \\ & = & \frac{1}{2}\end{array}\]

Our parabolic cross section has the equation

\[y(x)=\frac{1}{2}{x}^{2}\]

We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth \(y,\) the width will be given by \(2x,\) so we need to solve the equation above for \(x\) and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.

To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive \(x\) values. On this domain, we can find an inverse by solving for the input variable:

\[\begin{array}{lll}y & = & \frac{1}{2}{x}^{2} \\ 2y & = & {x}^{2} \\ x & = & \pm \sqrt{2y} \\ ∴y & = & \pm \sqrt{2x}\end{array}\]\[y=\sqrt{2x}\]\[\begin{array}{lll}\text{Area} & = & l⋅w \\ & = & 36⋅2x \\ & = & 72x \\ & = & 72\sqrt{2y}\end{array}\]
  1. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.
  2. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.
\[{f}^{-1}(f(x))=x,\ \text{for all }x\ \text{in the domain of }f\]

Condensed: the full section is in OpenStax College Algebra 2e.

Restricting the Domain to Find the Inverse of a Polynomial Function

So far, we have been able to find the inverse functions of cubic functions without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an inverse function. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses.

Example

Try it.

Find the inverse function of \(f\text{:}\)

  1. ⓐ \(f(x)={(x-4)}^{2},x\ge 4\)
  2. ⓑ \(f(x)={(x-4)}^{2},x\le 4\)
Solution

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.

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

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

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

Condensed: the full section is in OpenStax College Algebra 2e.

Key Concepts

  • The inverse of a quadratic function is a square root function.
  • If \({f}^{-1}\) is the inverse of a function \(f,\) then \(f\) is the inverse of the function \({f}^{-1}.\)
  • While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible.
  • To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one.
  • When finding the inverse of a radical function, we need a restriction on the domain of the answer.
  • Inverse and radical and functions can be used to solve application problems.

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Vježba (40)

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

    Otkrij odgovor

    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.

    Otkrij odgovor

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

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

    Otkrij odgovor

    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 \(f(x)=\sqrt[3]{x+4}.\)

    Otkrij odgovor

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

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

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

    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.

    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

    \[f(x)={(x-2)}^{2}-3.\]
    Otkrij odgovor

    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 \(x\ge 0.\)

    Otkrij odgovor

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

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

    Otkrij odgovor

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

    Otkrij odgovor

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

  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 \(\pi =3.14.\)

    Otkrij odgovor

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

    Otkrij odgovor

    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

    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.

    Otkrij odgovor

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

    Otkrij odgovor

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

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

    Otkrij odgovor

    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?

    Otkrij odgovor

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

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

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

    Otkrij odgovor

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

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

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

    Otkrij odgovor

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

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

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

    Otkrij odgovor

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

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

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

    Otkrij odgovor

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

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

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

    Otkrij odgovor

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

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

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

    Otkrij odgovor

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

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

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

    Otkrij odgovor

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

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

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

    Otkrij odgovor

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

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

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

    Otkrij odgovor

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

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

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

    Otkrij odgovor

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

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

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

    Otkrij odgovor

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

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

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

    Otkrij odgovor

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

Neka tvoja radi.

Besplatni račun dodaje bilješke na svaku lekciju, zapis onoga što ste završili, vaše riješene probleme na jednom mjestu, i tutora kojeg možete pitati o ovoj stranici. Sama matematika je otvorena za svakoga, prijavljenog ili ne.

Prijavite se Prijava

Simboli korišteni ovdje

Dodirnite bilo koji simbol za potpunu definiciju, sliku i što svako slovo u njemu znači.

Kako: Inverses and Radical Functions

  1. Find the inverse of a polynomial function.
  2. Restrict the domain to find the inverse of a polynomial function.
  3. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.
  4. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.
  5. Replace
  6. Interchange
  7. Solve for
  8. Restrict the domain by determining a domain on which the original function is one-to-one.

Pitanja koja ljudi postavljaju

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Dio ove stranice je prilagođen sa OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). U ovom slučaju, greške su naše. Budući da je izvor licenciran za nekomercijalnu upotrebu, ova lekcija je besplatna, ne prodaje ništa i dijeli se pod istom licencom.

Više u Precalculus

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