Answer:
[tex] \displaystyle 7[/tex]
Step-by-step explanation:
first thing I assume by f~¹ you meant [tex]f^{-1}[/tex] however
we want to find a²+3x-3 for the given condition. with the composite function condition we can do so
Finding the inverse of f(x):
[tex] \displaystyle f(x) = \sqrt{ax + 1} [/tex]
substitute y for f(x):
[tex] \displaystyle y= \sqrt{ax + 1} [/tex]
interchange:
[tex] \displaystyle x= \sqrt{ay + 1} [/tex]
square both sides:
[tex] \displaystyle ay + 1 = {x}^{2} [/tex]
cancel 1 from both sides:
[tex] \displaystyle ay = {x}^{2} - 1[/tex]
divide both sides by a:
[tex] \displaystyle y = \frac{{x}^{2} - 1 }{a}[/tex]
substitute f^-1 for y:
[tex] \displaystyle f ^{ - 1} (x) = \frac{{x}^{2} - 1 }{a}[/tex]
finding the inverse of g(x):
[tex] \displaystyle g(x) = \frac{x + 1}{x} [/tex]
substitute y for g(x)
[tex] \displaystyle y= \frac{x + 1}{x}[/tex]
interchange:
[tex] \displaystyle \frac{y + 1}{y} =x[/tex]
cross multiplication
[tex] \displaystyle y + 1= xy[/tex]
cancel 1 from both sides
[tex] \displaystyle y - xy= - 1[/tex]
factor out y:
[tex] \displaystyle y(1 - x)= - 1[/tex]
divide both sides by 1-x:
[tex] \displaystyle y= - \frac{1}{ 1 - x}[/tex]
substitute g^-1 for y:
[tex] \displaystyle g ^{ - 1} (x)= - \frac{1}{ 1 - x}[/tex]
remember that
[tex] \displaystyle (f \circ g)x = f(g(x))[/tex]
therefore we obtain:
[tex] \rm \displaystyle (f ^{ - 1} \circ g ^{ - 1} ) (3) = \frac{{ \bigg(- \dfrac{1}{1 - 3} } \bigg)^{2} - 1 }{a}[/tex]
since (f~¹•g~¹)(3)=-⅜ thus substitute:
[tex] \rm \displaystyle \frac{{ \bigg(- \dfrac{1}{1 - 3} } \bigg)^{2} - 1 }{a} = - \frac{3}{8} [/tex]
simplify parentheses:
[tex] \rm \displaystyle \frac{{ \bigg( \dfrac{1}{2} } \bigg)^{2} - 1 }{a} = - \frac{3}{8} [/tex]
simplify square:
[tex] \rm \displaystyle \frac{{ \dfrac{1}{4} } - 1 }{a} = - \frac{3}{8} [/tex]
simplify substraction:
[tex] \rm \displaystyle \frac{ - \dfrac{3}{4} }{ a}= - \frac{3}{8} [/tex]
simplify complex fraction:
[tex]\rm \displaystyle - \dfrac{3}{4a} = - \frac{3}{8}[/tex]
get rid of - sign:
[tex]\rm \displaystyle \dfrac{3}{4a} = \frac{3}{8}[/tex]
divide both sides by 3:
[tex]\rm \displaystyle \dfrac{1}{4a} = \frac{1}{8}[/tex]
cross multiplication:
[tex]\rm \displaystyle 4a= 8[/tex]
divide both sides by 4:
[tex]\rm \displaystyle \boxed{ a= 2}[/tex]
as we want to find a²+3a-3 substitute the got value of a:
[tex] \displaystyle {2}^{2} + 3.2 - 3[/tex]
simplify square:
[tex] \displaystyle 4 + 3.2 - 3[/tex]
simplify multiplication:
[tex] \displaystyle 4 +6 - 3[/tex]
simplify addition:
[tex] \displaystyle 10 - 3[/tex]
simplify substraction:
[tex] \displaystyle 7[/tex]
and we are done!
