Questions on Inverse Functions with Solutions

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.

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1) Graphical Solution: Locate points whose coordinates $(a , b)$ can easily be determined from the graph of $f$:
$$ (1 , 1), \quad (0 , -1), \quad (-1 , -3) $$

On the graph of the inverse function, these points will have swapped coordinates $(b , a)$:
$$ (1 , 1), \quad (-1 , 0), \quad (-3 , -1) $$

Plot the points and sketch the graph of the inverse of $f$ so that the two graphs are reflections across $y = x$.

2) Analytical Solution: Write the function as an equation in two variables:
$$ y = 2x^3 - 1 $$

Solve for $x$:
$$ 2x^3 = y + 1 $$
$$ x^3 = \frac{y + 1}{2} $$
$$ x = \sqrt[3]{\frac{y + 1}{2}} $$

Interchange $x$ and $y$ to write the inverse function:
$$ f^{-1} (x) = \sqrt[3]{\frac{x + 1}{2}} $$

Check: Verify the points $(-1, 0)$ and $(-3, -1)$ are on $f^{-1}$:
$$ f^{-1} (-1) = \sqrt[3]{\frac{-1 + 1}{2}} = 0 $$
$$ f^{-1} (-3) = \sqrt[3]{\frac{-3 + 1}{2}} = -1 $$
Both points work perfectly.

Step 1: Determine the Domain and Range mapping.
We are given a quadratic function with a restricted domain: $x \leq 2$. Let's write the function in vertex form by completing the square:
$$ f(x) = (x - 2)^2 + 1, \quad x \leq 2 $$

The vertex is $(2, 1)$ and the parabola opens upwards. Since the domain restricts us to the left half ($x \leq 2$), the range of $f$ is $y \geq 1$.
Therefore, for $f^{-1}$:
Domain of $f^{-1}$: $[1, \infty)$
Range of $f^{-1}$: $(-\infty, 2]$

Step 2: Write as an equation and solve for $x$.
$$ y = (x - 2)^2 + 1 $$
$$ (x - 2)^2 = y - 1 $$

Taking the square root gives two solutions:
$$ x - 2 = \pm \sqrt{y - 1} $$
$$ x = 2 + \sqrt{y - 1} \quad \text{or} \quad x = 2 - \sqrt{y - 1} $$

Step 3: Select the correct branch using the Range constraint.
Since the range of the inverse must be $x \leq 2$ (from the original domain), we must select the negative square root branch:
$$ x = 2 - \sqrt{y - 1} $$

Step 4: Interchange $x$ and $y$.
$$ f^{-1}(x) = 2 - \sqrt{x - 1} $$
Final Answer: $f^{-1}(x) = 2 - \sqrt{x - 1}$ with Domain $[1, \infty)$ and Range $(-\infty, 2]$.

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1) Graphical Solution: Locate a few points on the graph of $f(x)$. A possible list of coordinates $(a, b)$ is:
$$ (1.5, 0), \quad (2, 1), \quad (6, 3) $$

Swap coordinates to $(b, a)$ for the inverse graph:
$$ (0, 1.5), \quad (1, 2), \quad (3, 6) $$

Plot these points to sketch $f^{-1}$ reflecting over $y = x$.

2) Analytical Solution:
Domain/Range Setup: From the graph (or algebraically $2x-3 \geq 0$), the domain of $f$ is $x \geq 1.5$ and the range is $y \geq 0$. Therefore, the domain of $f^{-1}$ will be $x \geq 0$.

Write as an equation:
$$ y = \sqrt{2x - 3} $$

Square both sides and solve for $x$:
$$ y^2 = 2x - 3 $$
$$ 2x = y^2 + 3 $$
$$ x = \frac{y^2 + 3}{2} $$

Interchange $x$ and $y$ and apply the domain constraint:
$$ f^{-1}(x) = \frac{x^2 + 3}{2}, \quad x \geq 0 $$

Sketch the graph of $f^{-1}$ using the given graph of $y = f(x)$ and determine its equation analytically.

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1) Graphical Solution: Extract key vertices from the piecewise function $f(x)$:
$$ (0 , 3) , \quad (2 , -1) , \quad (5 , -3) $$

Swap coordinates to find vertices for $f^{-1}(x)$:
$$ (3 , 0) , \quad (-1 , 2) , \quad (-3 , 5) $$

2) Determining $f^{-1}(x)$ Analytically: We need to find the equations of the two linear segments for the inverse graph.

Segment 1: Between $x$-values $-3$ and $-1$. Passes through $(-3, 5)$ and $(-1, 2)$.
Slope $m_1$:
$$ m_1 = \frac{5 - 2}{-3 - (-1)} = \frac{3}{-2} = -\frac{3}{2} $$
Equation (using point-slope form with $(-1,2)$):
$$ y - 2 = -\frac{3}{2}(x - (-1)) \Rightarrow y = -\frac{3}{2}(x + 1) + 2 $$

Segment 2: Between $x$-values $-1$ and $3$. Passes through $(-1, 2)$ and $(3, 0)$.
Slope $m_2$:
$$ m_2 = \frac{0 - 2}{3 - (-1)} = \frac{-2}{4} = -\frac{1}{2} $$
Equation (using point-slope form with $(3,0)$):
$$ y - 0 = -\frac{1}{2}(x - 3) \Rightarrow y = -\frac{1}{2}(x - 3) $$
(Note: This is algebraically equivalent to $-\frac{1}{2}(x+1)+2$).

Conclusion (Piecewise Inverse):
$$ f^{-1}(x) = \begin{cases} -\frac{3}{2}(x + 1) + 2 & -3 \leq x \leq -1 \\ -\frac{1}{2}(x - 3) & -1 < x \leq 3 \end{cases} $$

Step 1: Determine Domain and Range.
$f(x)$ is defined if the radicand is $\geq 0$ and the denominator $\neq 0$:
$$ \frac{2}{x} - 1 \geq 0 \Rightarrow \frac{2 - x}{x} \geq 0 $$
Analyzing the signs at the critical values $x=0$ and $x=2$ gives the domain:
Domain of $f$: $(0, 2]$

Because the radical yields a positive number (or zero) and is multiplied by a negative sign, the range is all non-positive numbers.
Range of $f$: $(-\infty, 0]$

Therefore, for the inverse function:
Domain of $f^{-1}$: $(-\infty, 0]$
Range of $f^{-1}$: $(0, 2]$

Step 2: Sketch the Graph.
Points on $f$: $(2 , 0)$ and $(1 , -1)$.
Points on $f^{-1}$: $(0 , 2)$ and $(-1 , 1)$.

Step 3: Find $f^{-1}(x)$ Analytically.
$$ y = -\sqrt{\frac{2}{x} - 1} $$

Square both sides (this requires tracking our domain to avoid extraneous solutions):
$$ y^2 = \frac{2}{x} - 1 $$
$$ \frac{2}{x} = y^2 + 1 $$
$$ x = \frac{2}{y^2 + 1} $$

Interchange $x$ and $y$, and state the required domain restriction:
$$ f^{-1}(x) = \frac{2}{x^2 + 1}, \quad x \leq 0 $$

Below are shown the graphs of 6 functions. Sketch the graph of the inverse of each function.

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Solution: For each graph, select points whose coordinates are easy to determine. Use these points and the principle of reflection across the line $y = x$ to sketch the inverse functions.

Step 1: Determine the Domain and Range.
For the logarithm to be defined, the argument must be positive: $x + 2 > 0 \Rightarrow x > -2$.
Logarithms output all real numbers, so the range is unrestricted.
Domain of $f$: $(-2, \infty)$
Range of $f$: $(-\infty, \infty)$

Therefore, for the inverse function:
Domain of $f^{-1}$: $(-\infty, \infty)$
Range of $f^{-1}$: $(-2, \infty)$

Step 2: Solve algebraically.
$$ y = \log_4(x + 2) - 5 $$
$$ \log_4(x + 2) = y + 5 $$

Convert to exponential form:
$$ x + 2 = 4^{(y + 5)} $$
$$ x = 4^{(y + 5)} - 2 $$

Step 3: Interchange $x$ and $y$.
$$ f^{-1}(x) = 4^{(x + 5)} - 2 $$

If $f(x) = \ln(x) + 4x - 8$, what is the value of $f^{-1}(-4)$?

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Solution: We do not need to find the full inverse equation. Let $a = f^{-1}(-4)$. Using the compositional property of inverses, $f(f^{-1}(x)) = x$, we know that:
$$ f(a) = f(f^{-1}(-4)) = -4 $$

We need to find an $a$ such that $f(a) = -4$:
$$ \ln(a) + 4a - 8 = -4 $$
$$ \ln(a) = 4 - 4a $$

This transcendental equation cannot be solved purely algebraically, but its solution may be approximated graphically as the $x$-coordinate of the point of intersection of $y = \ln(x)$ and $y = 4 - 4x$.

The intersection is located exactly at $a = 1$. We can verify this algebraically:
$$ \ln(1) = 4 - 4(1) \Rightarrow 0 = 0 $$

Final Answer:
$$ f^{-1}(-4) = 1 $$