## Exercises - Linear Functions and Mobius Transformations

1. Find the linear function $f(x)$ whose graph has the given characteristics.

1. $m = \frac{2}{9}$, $y$-intercept $(0,4)$

2. $m = -\frac{8}{3}$, $y$-intercept $(0,-2)$

3. $m = -5$, $y$-intercept $(0,-\frac{2}{3})$

4. $m = -2$, passes through $(-5,1)$

5. $m = \frac{2}{3}$, passes through $(-4,-5)$

6. passes through $(-3,7)$ and $(-1,-5)$

7. $f(-5) = -3$ and $f(5) = 1$

See solutions with full work

1. $f(x) = \frac{2}{9} x + 4$
2. $f(x) = -\frac{8}{3} x - 2$
3. $f(x) = -5x-\frac{2}{3}$
4. $f(x) = -2x -9$
5. $f(x) = \frac{2}{3}x - \frac{7}{3}$
6. $f(x) = -6x-11$
7. $f(x) = \frac{2}{5} x - 1$
2. Determine if the following relations correspond to linear functions whose graphs are parallel or perpendicular (specify which), or something else.

1. $y = 3x + 1$ and $2y = 6x -7$

2. $y + 3x = 1$ and $y = \frac{1}{3} x + 1$

3. $2x + 5y = 4$ and $x = -\frac{5}{2} y - 7$

4. $y = 2x - 1$ and $y = -\frac{1}{2} x + 3$

5. $y = 7-x$ and $y = x+3$

6. $y + 3x = 2y - x$ and $y = 4x + 1$

See solutions with full work

1. parallel
2. perpendicular
3. parallel
4. perpendicular
5. perpendicular
6. parallel

3. Find the inverse of each function given, if it exists.

Note: the first three functions are linear, while the last three are Mobius transformations. Recall, when linear functions have an inverse, it will again be a linear function. Similarly, when a Mobius transformation has an inverse, it will again be a Mobius transformation.

1. $f(x) = -\frac{2}{3} x + 4$

2. $f(x) = 6 x - \frac{1}{2}$

3. $f(x) = 7$

4. $f(x) = \cfrac{x+4}{x-3}$

5. $f(x) = \cfrac{5x-3}{2x+1}$

6. $f(x) = \cfrac{x+6}{3x-4}$

See solutions with full work

1. $f^{-1}(x) = -\frac{3}{2} x + 6$

2. $f^{-1}(x) = \frac{1}{6} x + \frac{1}{12}$

3. no inverse exists -- $f$ graphs as a horizontal line which spectacularly fails the horizontal line test

4. $f^{-1}(x) = \cfrac{3x+4}{x-1}$

5. $f^{-1}(x) = \cfrac{-x-3}{2x-5}$

6. $f^{-1}(x) = \cfrac{4x+6}{3x-1}$

4. Find the inverse of $f(x) = \cfrac{6x + 8}{2x+1}$ in two different ways.

For one of the ways, do the division first, following this with a "socks-and-shoes" argument.
5. For each pair of functions, find $(f \circ g)(x)$.

Note: each composition sought is either a composition of linear functions (which must then be linear), or a composition of Mobius transformations (which then must be another Mobius transformation).

1. $f(x) = 2x - 1$; $g(x) = -\frac{1}{3}x + 5$

2. $f(x) = -\frac{3}{7}x + 2$; $g(x) = \frac{1}{2}x + \frac{2}{3}$

3. $\displaystyle{f(x) = \frac{2x-1}{3x+5}}$; $\displaystyle{g(x) = \frac{7x+1}{3x-4}}$

4. $f(x) = 3x+5$; $\displaystyle{g(x) = \frac{1}{x+2}}$

Don't worry -- these last two functions really are both Mobius transformations, as $\frac{1}{x+2} = \frac{0x+1}{1x+2}$ and $3x+5 = \frac{3x+5}{0x+1}$. Indeed, as this last calculation suggests, every linear function is also a Mobius transformation -- just not always the other way around.