.
1) Sketch the graph of the inverse of \( f \) in the same system of axes.
2) Find the inverse of \( f \) and check your answer using some points.
Learn how to solve math problems involving inverse functions using both analytical techniques and graphing methods. This page offers step-by-step solutions to a variety of questions, highlighting key concepts such as the reflection property: the graph of a function and its inverse are mirror images across the line \( y = x \). Ideal for students and educators seeking clear explanations and visual understanding of inverse functions.
Below is shown the graph of \[ f(x) = 2 x^3 - 1 \]
.
1) Sketch the graph of the inverse of \( f \) in the same system of axes.
2) Find the inverse of \( f \) and check your answer using some points.
1) Locate Points on the Graph of \( f \). Here is a list of points whose coordinates \( (a , b) \) can easily be determined from the graph: \[ (1 , 1), \quad (0 , -1), \quad (-1 , -3) \]
On the graph of the inverse function, the above points will have coordinates \( (b , a) \) as follows: \[ (1 , 1), \quad (-1 , 0), \quad (-3 , -1) \]
Plot the above points and sketch the graph of the inverse of \( f \) so that the two graphs are reflections of each other on the line \( y = x \) as shown below.
2) Write the given function \( f(x) = 2x^3 - 1 \) as an equation in two unknowns: \[ y = 2x^3 - 1 \]
Solve the above for \( x \): \[ 2x^3 = y + 1 \] \[ x^3 = \dfrac{y + 1}{2} \] \[ x = \sqrt[3]{\dfrac{y + 1}{2}} \]
Interchange x and y and write the equation of inverse function \( f^{-1} \): \[ y = \sqrt[3]{\dfrac{x + 1}{2}} \] \[ f^{-1} (x) = \sqrt[3]{\dfrac{x + 1}{2}} \]
We now verify that the points \( \quad (1 , 1), \quad (-1 , 0), \quad (-3 , -1) \quad \) used above to sketch the graph of the inverse function are on the graph of \( f^{-1} \). \[ f^{-1} (1) = \sqrt[3]{\dfrac{1 + 1}{2}} = 1 \] \[ f^{-1} (-1) = \sqrt[3]{\dfrac{-1 + 1}{2}} = 0 \] \[ f^{-1} (-3) = \sqrt[3]{\dfrac{-3 + 1}{2}} = -1 \]
Let \( f(x) = x^2 - 4x + 5 \), \( x \leq 2 \).
1) Find the inverse function of \( f \).
2) Find the domain and the range of \( f^{-1} \).
1) We are given a quadratic function with a restricted domain. We first write the given function in vertex form (may be done by completing the square):
\[ f(x) = x^2 - 4x + 5 = (x - 2)^2 + 1, \quad x \leq 2 \]
The graph of function \( f \) is that of the left half of a parabola with vertex at \( (2, 1) \) as shown below.
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We now write the given function as an equation. \[ y = (x - 2)^2 + 1 \]
Solve the above for \( x \) \[ y = (x - 2)^2 + 1 \] \[ (x - 2)^2 = y - 1 \]
Two solutions: \[ x - 2 = \pm \sqrt{y - 1} \] which gives \[ x = \sqrt{y - 1} + 2 \quad \text{and} \quad x = -\sqrt{y - 1} + 2 \]
Since \( x \leq 2 \) (domain of \( f \)) and \( \sqrt{y - 1} \ge 0 \), we select the solution
\[ x = -\sqrt{y - 1} + 2 \]
Interchange \( x \) and \( y \) to write the inverse of function \( f \) as follows. \[ y = f^{-1}(x) = -\sqrt{x - 1} + 2 \] The domain and range of \( f^{-1} \) are the range and domain of \( f \).
Domain of \( f^{-1} \) is the range of \( f \): \[ [1, +\infty) \quad \text{from graph} \] Range of \( f^{-1} \) is the domain of \( f \): \[ (-\infty, 2] \quad {given} \]
Below is shown the graph of \[ f(x) = \sqrt{2 x - 3} \]
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1) Sketch the inverse of \( f \) in the same graph.
2) Find the inverse of and check your answer using some points.
1) Locate a few points on the graph of \( f \). A possible list of points with coordinates \( (a, b) \) is as follows: \[ (1.5, 0) , (2, 1) , (6, 3) \]
On the graph of the inverse function, the above points will have coordinates \( (b, a) \) as follows: \[ (0, 1.5) , (1, 2) , (3, 6) \]
Plot the above points and sketch the graph of the inverse of \( f \) so that the two graphs are reflections of each other on the line \( y = x \) as shown below.
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2) Write the given function \( f(x) = \sqrt{2x - 3} \) as an equation in two unknowns. \[ y = \sqrt{2x - 3} \]
Solve the above for \( x \). First square both sides: \[ 2x - 3 = y^2 \] \[ 2x = y^2 + 3 \] Solve \[ x = \dfrac{y^2 + 3}{2} \]
Interchange \( x \) and \( y \) and write the equation of the inverse function \( f^{-1} \); and write the domain of the inverse. \[ y = \dfrac{x^2 + 3}{2} \] The domain which of \( f^{-1} \) is the range of \( f \) from its graph above, hence \[ f^{-1}(x) = \dfrac{x^2 + 3}{2}, \quad x \geq 0 \] We now verify that the points \( (0 , 1.5) \), \( (1 , 2) \), and \( (3 , 6) \) used to sketch the graph of the inverse function are on the graph of \( f^{-1} \). \[ f^{-1}(0) = \dfrac{0^2 + 3}{2} = 1.5 \] \[ f^{-1}(1) = \dfrac{1^2 + 3}{2} = 2 \] \[ f^{-1}(3) = \dfrac{3^2 + 3}{2} = 6 \]
Sketch the graph of \( f^{-1} \) using the graph of \( y = f(x) \) shown below and find \( f^{-1}(x) \).
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1) Use the graph to find points on the graph of \( f \). A possible list of points whose coordinates \( (a , b) \) is as follows: \[ (0 , 3) , (2 , -1) , (5 , -3) \]
On the graph of the inverse function, the above points will have coordinates \( (b , a) \) as follows: \[ (3 , 0) , (-1 , 2) , (-3 , 5) \]
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2) Determining the inverse function \( f^{-1}(x) \)
For \( -3 \leq x \leq -1 \), \( f^{-1}(x) \) is a linear expression with slope \( m_1 \), passing through the points \( (-1, 2) \) and \( (-3, 5) \):
\[ m_1 = \dfrac{5 - 2}{-3 - (-1)} = \dfrac{-3}{2} \]Therefore, for \( -3 \leq x \leq -1 \), the inverse function is:
\[ f^{-1}(x) = -\dfrac{3}{2}(x - (-1)) + 2 = -\dfrac{3}{2}(x + 1) + 2 \]For \( -1 < x \leq 3 \), \( f^{-1}(x) \) is a linear expression with slope \( m_2 \), passing through the points \( (-1, 2) \) and \( (3, 0) \):
\[ m_2 = \dfrac{0 - 2}{3 - (-1)} = \dfrac{-2}{4} = -\dfrac{1}{2} \]Therefore, for \( -1 < x \leq 3 \), the inverse function is:
\[ f^{-1}(x) = -\dfrac{1}{2}(x - (-1)) + 2 = -\dfrac{1}{2}(x + 1) + 2 \] Conclusion: \[ f^{-1}(x) = \left \{ \begin{array}{ll} -\dfrac{3}{2}(x + 1) + 2 \quad -3 \leq x \leq -1 \\ -\dfrac{1}{2}(x + 1) + 2 \quad -1 \lt x \leq 3 \end{array} \right \} \] The one to one function
\[ f(x) = - \sqrt{\dfrac{2}{x}-1} \]
is graphed below.
1) What is the domain and range of \( f \)?
2) Sketch the graph of \( f^{-1} \).
3) Find \( f^{-1}(x) \) (include domain).
1) \( f(x) \) is defined as a real number if the radicand \( \dfrac{2}{x} - 1 \) is greater than or equal to 0. Hence, we need to solve the inequality:
\[
\dfrac{2}{x} - 1 \geq 0
\]
\[
\dfrac{2 - x}{x} \geq 0
\]
The expression on the left of the inequality changes sign at the zeros of the numerator and denominator, which are \( x = 2 \) and \( x = 0 \). See table below.
.
Domain of \( f \): \( (0 , 2] \)
Range of \( f \): \( (-\infty , 0] \)
2) Points on the graph of \( f \) \[ (2 , 0) , (1 , -1) \] The above points on the graph of the inverse function will have coordinates \( (b , a) \) as follows: \[ (0 , 2) , (-1 , 1) \]
Plot the above points and sketch the graph of the inverse of \( f \) so that the two graphs are reflection of each other on the line \( y = x \) as shown below.
.
3) Write \( f(x) \) as an equation in \( y \) and \( x \). \[ y = -\sqrt{\dfrac{2}{x} - 1} \]
Solve the above equation for \( x \). Square both sides of the above equation: \[ y^2 = \dfrac{2}{x} - 1 \] \[ \dfrac{2}{x} = y^2 + 1 \] \[ x = \dfrac{2}{y^2 + 1} \]
Interchange \( x \) and \( y \) and write the inverse function: \[ y = \dfrac{2}{x^2 + 1} \] \[ f^{-1}(x) = \dfrac{2}{x^2 + 1} \]
Domain and range of \( f^{-1} \) are the range and domain of \( f \). Hence:
Domain of \( f^{-1} \): \[ (-\infty , 0] \] Range of \( f^{-1} \): \[ (0 , 2] \]
.
.
Find the inverse of \[ f(x) = \log_4 (x + 2) - 5 \] , its domain and range.
Write the given function as an equation in \( x \) and \( y \) as follows: \[ y = \log_{4}(x + 2) - 5 \]
Solve the above equation for \( x \). \[ \log_{4}(x + 2) = y + 5 \] \[ x + 2 = 4^{(y + 5)} \] \[ x = 4^{(y + 5)} - 2 \]
Interchange \( x \) and \( y \). \[ y = 4^{(x + 5)} - 2 \]
Write the inverse function with its domain and range. \[ f^{-1}(x) = 4^{(x + 5)} - 2, \quad \text{Domain: } (-\infty, +\infty), \quad \text{Range: } (-2, +\infty) \]
If \( f(x) = \ln(x) + 4x - 8 \), what is the value of \( f^{-1}(-4) \)?
Let \( a = f^{-1}(-4) \). Then Using the property \( f(f^{-1}(x)) = x \) of the inverse function, we write: \[ f(a) = f(f^{-1}(-4)) = -4 \]
We now need to find \( a \) such that \( f(a) = -4 \), hence the equation to solve:
\[ \ln(a) + 4a - 8 = -4 \] \[ \ln(a) = 4 - 4a \]
The above equation cannot be solved analytically but its solution may be approximated graphically as the \( x \)-coordinate of the point of intersection of the graphs of \( y = \ln(x) \) and \( y = 4 - 4x \), as shown below.
The intersection of the two graphs is close to \( x = 1 \), which can easily be checked to be the exact solution to the equation \( \ln(x) = 4 - 4x \). Hence, \[ f^{-1}(-4) = 1 \]