Inverse equation: $f^{- 1} (x) = \frac{- 11 x - 1}{- 7 x + 3}$ $f^-1(x)=(-11x-1)/(-7x+3)$ were $x \neq \frac{3}{7}$ $x!=3/7$ , what would the domain be?

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Inverse equation: $f^{- 1} (x) = \frac{- 11 x - 1}{- 7 x + 3}$ $f^-1(x)=(-11x-1)/(-7x+3)$ were $x \neq \frac{3}{7}$ $x!=3/7$ , what would the domain be?

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Answer 1 · 2017-06-06T14:18:28

$D_f=x in RR, x!=3/7$

Explanation:

$f^-1(x)=(-11x-1)/(-7x+3)$

Apart from $x!=3/7$ , there are no other values for which $f^-1$ is undefined, thus $x$ can take all real values barring $3"/"7$ .
graph{(-11x-1)/(-7x+3) [-10, 10, -5, 5]}

You can see this from the graph where $f^-1$ has only one asymptote at $x=3"/"7$

Answer 2 · 2017-06-06T14:28:54

$x inRR,x!=3/7$

Explanation:

$" the domain is all real values apart from any that make the"$
$"denominator zero which make the function undefined"$

$"equating the denominator to zero and solving gives the "$
$"value that x cannot be"$

$"solve " -7x+3=0rArrx=3/7larrcolor(red)" excluded value"$

$rArr"domain is " x inRR,x!=3/7$

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Inverse equation: $f^{- 1} (x) = \frac{- 11 x - 1}{- 7 x + 3}$ $f^-1(x)=(-11x-1)/(-7x+3)$ were $x \neq \frac{3}{7}$ $x!=3/7$ , what would the domain be?

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Explanation:

Explanation:

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Inverse equation: f^-1(x)=(-11x-1)/(-7x+3)f−1(x)=−11x−1−7x+3f^-1(x)=(-11x-1)/(-7x+3) were x!=3/7x≠37x!=3/7, what would the domain be?

2 Answers

Explanation:

Explanation:

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Inverse equation: $f^{- 1} (x) = \frac{- 11 x - 1}{- 7 x + 3}$ $f^-1(x)=(-11x-1)/(-7x+3)$ were $x \neq \frac{3}{7}$ $x!=3/7$ , what would the domain be?