# Let f(x)=5x+4 and g(x)=x−4/5 , find: a). (f@g)(x) ? b). (g@f)(x) ?

Feb 26, 2018

(f ∘ g) (x) = 5x (g ∘ f) (x)=5x+16/5

#### Explanation:

Finding (f ∘ g) (x) means finding $f \left(x\right)$ when it is composed with $g \left(x\right)$, or $f \left(g \left(x\right)\right)$ . This means replacing all instances of $x$ in $f \left(x\right) = 5 x + 4$ with $g \left(x\right) = x - \frac{4}{5}$:

(f ∘ g) (x)=5(g(x))+4=5(x-4/5)+4=5x-4+4=5x

Thus, (f ∘ g) (x) = 5x

Finding (g ∘ f) (x) means finding $g \left(x\right)$ when it is composed with $f \left(x\right)$, or $g \left(f \left(x\right)\right) .$ This means replacing all instances of $x$ in $g \left(x\right) = x - \frac{4}{5}$ with $f \left(x\right) = 5 x + 4 :$

(g ∘ f) (x) = f(x)-4/5=5x+4-4/5=5x+20/5-4/5=5x+16/5

Thus, (g ∘ f) (x)=5x+16/5

Feb 26, 2018

See explanation...

#### Explanation:

Alright, first lets remember what $f \circ g$ and $g \circ f$ mean.

$f \circ g$ is a fancy way of saying $f \left(g \left(x\right)\right)$ and $g \circ f$ is a fancy way of saying $g \left(f \left(x\right)\right)$. Once we realize this, these problems aren't that difficult to solve.

So $f \left(x\right) = 5 x + 4$ and $g \left(x\right) = x - \frac{4}{5}$

a) $f \circ g$

Ok lets start with the $f \left(x\right)$ function

$f \left(x\right) = 5 x + 4$

Then, we just add the $g \left(x\right)$ function whenever we see an $x$ in the $f \left(x\right)$ function.

$f \left(g \left(x\right)\right) = 5 g \left(x\right) + 4$$\to$$5 \left(x - \frac{4}{5}\right) + 4$

Simplify:

f(g(x)))=(5x-4)+4 $\to$ $5 x \cancel{- 4} \cancel{+ 4}$

So therefore, $f \circ g = 5 x$

b) $g \circ f$

Alright, it's the same process here just it's the opposite. Let's start with the $g \left(x\right)$ function.

$g \left(x\right) = x - \frac{4}{5}$

Then, we just add the $f \left(x\right)$ function whenever we see an $x$ in the $g \left(x\right)$ function.

$g \left(f \left(x\right)\right) = f \left(x\right) - \frac{4}{5}$$\to$$\left(5 x + 4\right) - \frac{4}{5}$

Simplify:

$g \left(f \left(x\right)\right) = 5 x + \frac{16}{5}$

Therefore, $g \circ f = 5 x + \frac{16}{5}$

Hope this helped!
~Chandler Dowd

Mar 22, 2018

For $g \left(x\right) = x - \frac{4}{5}$ it is solved by Chandler Dowd and VNVDVI
For $g \left(x\right) = \frac{x - 4}{5}$, [ requested by Widi K. ]the solution is

$\textcolor{red}{\left(f o g\right) \left(x\right) = x \mathmr{and} \left(g o f\right) \left(x\right) = x}$

#### Explanation:

We have,f(x)=color(red)(5x+4 ...to(1)
and g(x)=color(blue)((x-4)/5.......to(2).
Hence,
$\left(f o g\right) \left(x\right) = f \left(g \left(x\right)\right)$
$\left(f o g\right) \left(x\right) = f \left(\textcolor{b l u e}{\frac{x - 4}{5}}\right) \ldots . \to$from(2)
$\left(f o g\right) \left(x\right) = f \left(m\right)$,......[ take $m = \frac{x - 4}{5}$ ]
(fog)(x)=color(red)(5m+4......[Apply (1) for $x \to m$]
$\left(f o g\right) \left(x\right) = \cancel{5} \left(\textcolor{b l u e}{\frac{x - 4}{\cancel{5}}}\right) + 4$...[ put $m = \frac{x - 4}{5}$ ]
$\left(f o g\right) \left(x\right) = x - 4 + 4$
$\left(f o g\right) \left(x\right) = x$

$\left(g o f\right) \left(x\right) = g \left(f \left(x\right)\right)$
$\left(g o f\right) \left(x\right) = g \left(\textcolor{red}{5 x + 4}\right) \ldots \ldots \to$from(1)
$\left(g o f\right) \left(x\right) = g \left(n\right) \ldots \ldots . .$[ take $n = 5 x + 4$
$\left(g o f\right) \left(x\right) = \left(\textcolor{b l u e}{\frac{n - 4}{5}}\right)$......[ Apply (2) for $x \to n$
$\left(g o f\right) \left(x\right) = \frac{5 x + 4 - 4}{5.} \ldots$[ put n=5x+4 ]
$\left(g o f\right) \left(x\right) = \frac{5 x}{5}$
$\left(g o f\right) \left(x\right) = x$