Question

Ayudenme Por Favor Con esto Ejercicio

Answer 1

$a.\:f\left(x\right)=4x+1,\:\:g\left(x\right)=3x,\:\:(f\circ \:g)_{x}$
$\mathrm{Para}\:f=4x+1\:\mathrm{sustituir}\:x\:\mathrm{con}\:g\left(x\right)=3x:$
$(f\circ \:g)_{x}=4\cdot \:(3x)+1$
$(f\circ \:g)_{x}=12x+1$

$f\left(x\right)=4x+1,\:\:g\left(x\right)=3x,\:(g\circ \:f)_{x}$
$\mathrm{Para}\:g=3x\:\mathrm{sustituir}\:x\:\mathrm{con}\:f\left(x\right)=4x+1:$
$(g\circ \:f)_{x}=3\left(4x+1\right)$
$\mathrm{Poner\:los\:parentesis\:utilizando}:\quad \:a\left(b+c\right)=ab+ac$
$(g\circ \:f)_{x}=12x+3$

$b.\:f\left(x\right)=\sqrt{x-5},\:\:g\left(x\right)=\frac{x}{5},\:\:(f\circ \:g)_{x}$
$\mathrm{Para}\:f=\sqrt{x-5}\:\mathrm{sustituir}\:x\:\mathrm{con}\:g\left(x\right)=\frac{x}{5}:$
$(f\circ \:g)_{x}=\sqrt{\frac{x}{5}-5}$

$f\left(x\right)=\sqrt{x-5},\:\:g\left(x\right)=\frac{x}{5},\:\:(g\circ \:f)_x$
$\mathrm{Para}\:g=\frac{x}{5}\:\mathrm{sustituir}\:x\:\mathrm{con}\:f\left(x\right)=\sqrt{x-5}:$
$(g\circ \:f)_x=\frac{\sqrt{x-5}}{5}$