The function $f$ is defined by $f:x\mapsto x-2$ and another function $g$ is such that
$$g\circ f:x\mapsto \frac{1}{11-3x} , x≠\frac{11}{3}$$
Find the function $g$.
f is defined by $f:x\mapsto x-2$ and g is a function such that $g\circ f:x\mapsto\frac{1}{11-3x}$ , $x≠ \frac{11}{3}$ Find the function g
0
$\begingroup$
functions
function-and-relation-composition
-
0Is $gf$ compsoition or product? – 2017-01-15
-
0composition, i think – 2017-01-15
-
0Yeah, maybe its composition otherwise the solution is trivial. – 2017-01-15
-
0You know that $(g \circ f)(x)=1 / (11-3x) = h(x)\,$. Then $g = h \circ f^{-1}$ where $f^{-1}$ is easy to determine. – 2017-01-15
2 Answers
2
$$ f(x) = x-2 $$ $$ g(f(x)) = \frac{1}{11-3x} $$ $$ g(x-2) = \frac{1}{11-3x} $$ Substitute $t = x-2$ $$ g(t) = \frac{1}{11-3(t+2)} = \frac{1}{5-3t} $$
0
If you mean $gf$ as a composition, then $g(x)=\frac{1}{5-3x}$.