Please answer quickly

Part 1. Using the two functions listed below, insert numbers in place of the letters a, b, c, and d so that f(x) and g(x) are inverses.

f(x)=x+a/b

g(x)=cx-d


Part 2. Show your work to prove that the inverse of f(x) is g(x).

Part 3. Show your work to evaluate g(f(x)). Part 4. Graph your two functions on a coordinate plane. Include a table of values for each function. Include five values for each function. Graph the line y=x on the same graph.

Part 4. Graph your two functions on a coordinate plane. Include a table of values for each function. Include five values for each function. Graph the line y=x on the same graph.

Task 2

Part 1. Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation

a√x+b+c=d

Part 2. Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation. Part 2. Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneous.

Part 3. Explain why the first equation has an extraneous solution and the second does not.

f(x)=(x+a)/b

or bf(x)=x+a

let f(x)=y

by=x+a

flip x and y

bx=y+a

or y=bx-a

or f^{-1}(x)=bx-a

also g(x) is inverse of f(x)

bx-a=cx-d

so b=c,a=d

again let g(x)=y

y=cx-d

flip x and y

x=cy-d

cy=x+d

y=(x+d)/c

or g^{-1}(x)=(x+d)/c

also f(x) is inverse of g(x)

so (x+a)/b=(x+d)/c

so a=d,b=c

so in either case a=d,b=c

take b=c=1

a=d=2

f(x)=(x+2)/1=x+2

g(x)=1x-2=x-2

so f(x) and g(x) are two parallel lines f(x) with y- intercept=1 and slope 0

g(x) with y-intercept -2 and slope 0

if we take b=c=2,a=d=3

f(x)=(x+3)/2=x/2+3/2

g(x)=2x-3

here f(x) is of slope 1/2 and y-intercept 3/2

g(x) is of slope 2 and y intercept -3

part 3.

f(f(x))=g((x+a)/b)=c[(x+a)/b]-d=(c/b)(x+a)-d

An inverse function or an anti function exists described as a function, which can change into another function. In other words, if any function “f” carries x to y then, the inverse of “f” will carry y to x.

Part 1: f(x) = (x+2) / 3 and g(x) = 3y-2.

Part 2: g(x) = 3x - 2, the inverse.

Part 3: g(f(x)) = x showing that f(x) and g(x) are mutual inverses.

An inverse function in mathematics exists function which "reverses" the another function.

TASK 1

Part 1

Let y = f(x) = (x + a) / b

x = g(y) = by-a = cy-d,

so c = b and d = a.

Let d = a = 2 and c = b = 3

f(x) = (x+2) / 3 and g(x) = 3y-2.

Part 2

y = f(x) = (x+2) / 3

3y = x+2  

x = 3y-2 = g(y)

so g(x) = 3x - 2, the inverse.

Part 3

g(f(x)) = 3

f(x)-2 = 3(x+2)/3-2 = x+2-2 = x,

so g(f(x)) = x showing that f(x) and g(x) exists mutual inverses.

x       -2  -1        0    1     2

f(x)    0    1/3   2/3   1    4/3

g(x)  -8   -5      -2    1      4

Part 4

The graph is given below.

TASK 2

Part 1

(a) Let a=1, b=2, c=3, d=6: [tex]$\sqrt{x}[/tex]+2+3=6 ; [tex]\sqrt{x}[/tex]-6=-5

(b) Let a=-1, b=2, c=3, d=4

(a) Multiply both sides by

[tex]$\sqrt{x}+6: x-36=-5(\sqrt{x}+6)=-5 \sqrt{x}-30[/tex]

[tex]$ x-36+30=-5 \sqrt{x }[/tex]

[tex]$ x-6=-5 \sqrt{x}$[/tex]

Square both sides:

[tex]$x^{2} -12 x+36=25 x ; x^{2} -37 x+36=0=(x-36)(x-1)$[/tex].

Part 2

(a) So x=36 or 1 (apparently).

Substitute x=36 in the original equation: 6+2+3=6, 11=6 is not true.

So x=36 is an extraneous solution.

substitute x=1: 1+2+3=6 is true, so x=1 exists the actual solution.

(b) [tex]$-\sqrt{x}+2+3=4 ;-\sqrt{x}=-1$[/tex]; multiply both sides by -1: [tex]$\sqrt{x}=1$[/tex]; square both sides: x=1.

Substitute x=1 in the original equation: -1+2+3=4 is true.

So x = 1 is the solution.

Part 3

(a) the act of squaring both sides resulted in creating an extraneous solution [tex]$\left((-5)^{2} =25=5^{2} \right)$[/tex].  

(b) the solution was simpler and there was no ambiguity.

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