Question #d780a

Question

1225 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-05T08:49:24

${(g \circ f)}^{- 1} = - \frac{1}{28} (x + 11)$ $(g circ f)^(- 1) = - frac(1)(28) (x + 11)$

We have: $f (x) = 4 x + 3$ $f(x) = 4 x + 3$ and $g (x) = - 7 x + 10$ $g(x) = - 7 x + 10$

First, let's evaluate $g \circ f$ $g circ f$ :

$\Rightarrow g \circ f = g (f (x))$ $Rightarrow g circ f = g (f(x))$

$\Rightarrow g \circ f = g (4 x + 3)$ $Rightarrow g circ f = g (4 x + 3)$

$\Rightarrow g \circ f = - 7 (4 x + 3) + 10$ $Rightarrow g circ f = - 7 (4 x + 3) + 10$

$\Rightarrow g \circ f = - 28 x - 21 + 10$ $Rightarrow g circ f = - 28 x - 21 + 10$

$\Rightarrow g \circ f = - 28 x - 11$ $Rightarrow g circ f = - 28 x - 11$

Let's suppose $y = g \circ f$ $y = g circ f$ .

We must now find the inverse of $y$ $y$ :

$\Rightarrow y = - 28 x - 11$ $Rightarrow y = - 28 x - 11$

Interchanging variables:

$\Rightarrow x = - 28 y - 11$ $Rightarrow x = - 28 y - 11$

Let's solve for $y$ $y$ :

$\Rightarrow x + 11 = - 28 y$ $Rightarrow x + 11 = - 28 y$

$\Rightarrow y = - \frac{1}{28} (x + 11)$ $Rightarrow y = - frac(1)(28) (x + 11)$

$therefore (g circ f)^(- 1) = - frac(1)(28) (x + 11)$