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Find the linear function $f(x)$ whose graph has the given characteristics.
$m = \frac{2}{9}$, $y$-intercept $(0,4)$
$m = -\frac{8}{3}$, $y$-intercept $(0,-2)$
$m = -5$, $y$-intercept $(0,-\frac{2}{3})$
$m = -2$, passes through $(-5,1)$
$m = \frac{2}{3}$, passes through $(-4,-5)$
passes through $(-3,7)$ and $(-1,-5)$
$f(-5) = -3$ and $f(5) = 1$
Determine if the following relations correspond to linear functions whose graphs are parallel or perpendicular (specify which), or something else.
$y = 3x + 1$ and $2y = 6x -7$
$y + 3x = 1$ and $y = \frac{1}{3} x + 1$
$2x + 5y = 4$ and $x = -\frac{5}{2} y - 7$
$y = 2x - 1$ and $y = -\frac{1}{2} x + 3$
$y = 7-x$ and $y = x+3$
$y + 3x = 2y - x$ and $y = 4x + 1$
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.
$f(x) = -\frac{2}{3} x + 4$
$f(x) = 6 x - \frac{1}{2}$
$f(x) = 7$
$f(x) = \cfrac{x+4}{x-3}$
$f(x) = \cfrac{5x-3}{2x+1}$
$f(x) = \cfrac{x+6}{3x-4}$
$f^{-1}(x) = -\frac{3}{2} x + 6$
$f^{-1}(x) = \frac{1}{6} x + \frac{1}{12}$
no inverse exists -- $f$ graphs as a horizontal line which spectacularly fails the horizontal line test
$f^{-1}(x) = \cfrac{3x+4}{x-1}$
$f^{-1}(x) = \cfrac{-x-3}{2x-5}$
$f^{-1}(x) = \cfrac{4x+6}{3x-1}$
Find the inverse of $f(x) = \cfrac{6x + 8}{2x+1}$ in two different ways.
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).
$f(x) = 2x - 1$; $g(x) = -\frac{1}{3}x + 5$
$f(x) = -\frac{3}{7}x + 2$; $g(x) = \frac{1}{2}x + \frac{2}{3}$
$\displaystyle{f(x) = \frac{2x-1}{3x+5}}$; $\displaystyle{g(x) = \frac{7x+1}{3x-4}}$
$f(x) = 3x+5$; $\displaystyle{g(x) = \frac{1}{x+2}}$
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.