Find the inverse function of f(x)= 1/3x−1 . Let, f(x)=3x+7,g(x)=2√x−3 . Find f(g(3)) and g(f(3)).

Answer 1

the problem is a little ambiguous

IF you mean y= (1/3)x -1

switch x and y, solve for y

x = y/3 -1

y/3 = x+1

y = 3x +3

f^-1(x) = 3x +3 is the inverse function

UNLESS you really meant y=1/(3x) - 1

then

x = 1/3y -1

x+1 = 1/3y

3y = 1/(x+1)

y =1/(3x+3)

f^-1(x) = (3x+3)^-1 is the inverse function

(1/3x using PEMDAS means 1/(3x), not (1/3)x, as M comes before D in PEMDAS, so technically the 2nd solution is correct, but odds are you really meant the first problem)

2nd problem has a similar ambiguity as to the square root sign. Did you mean sqr(x) -3 or sqr(x-3)?

the way it's written, you probably mean sqr(x) - 3

IF so, then

g(3) = 2sqr3 - 3

and

f(g(3)) = 3(2sqr3 -3) + 7 = 6sqr3 -2

f(3) = 3(3) +7 = 16

g(f(3)) = 2sqr16 - 3 = 2(4)-3 = 5

But if you did intend the square root sign to encompass both terms

then

g(3) = 2sqr(3-3) = 0

f(0) = 3(0) + 7 = 7

f(3) = 3(3)+7 = 16

g(16) = 2sqr(16-3) = 2sqr13

Answer 2

Find the inverse function of f(x)= 1/3x−1

f(x)=y=1/3x-1

Replace y by x and x by y

x=1/3y-1

Solve for y

x+1=1/3y

y=3x+3 is the inverse

f(x)=3x+7,g(x)=2√x−3 . Find f(g(3)) and g(f(3)).

g(x)=2√x−3

g(3)=2√3−3 (Replace x=3 in the given g(x))

f(g(3))=f(2√3−3) = 3(2√3−3)+7 =6√3-2

f(x)=3x+7

f(3)= 3(3)+7=16 (Replace x=3 in the given f(x))

g(f(3))=g(16)=2√16−3= 2(4)-3=5