Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

# 6. Suppose g(x) = 3+(x/2x-3).  Evaluate g^-1(-1)?

Precalc

Relevance

Given: g(x) = 3 + (x / (2x - 3))

Find the inverse, switch x and y and solve for y:

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

x - 3 = y / (2y - 3)

(2y - 3)(x - 3) = y

2xy - 3x - 6y + 9 - y = 0

2xy - 6y - y = 3x - 9

y(2x - 7) = 3x - 9

y = (3x - 9) / (2x - 7)

g^-1(-1) = (3(-1) - 9) / (2(-1) - 7)

g^-1(-1) = 4/3

• Let g(x0 = y

Hence

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

Subtract '3' from both sides

y - 3 = x/(2x - 3)

y(2x - 3) = x

2xy - 3y = x

2xy - x = 3y

x(2y - 1) = 3y

x = 3y / (2y - 1)

'Swop' the letters

y = 3x/(2x - 1)

g^-1(x) = 3x / (2x - 1)

g^-1(-1) = 3(-1) / (2(-1) - 1)

g^-1(-1) = -3 / ( - 2 - 1)

g^-1(-1) = -3 / -3  = 3/3 = 1

Hence g^-1(-1) = 1

• (3 + x) / 2x

• g(x) = 3 + x / ( 2x - 3 )

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

( y - 3 ) =  x / ( 2x - 3 )

( y - 3 )( 2x - 3 ) = x

2xy - 6x - 3y + 9 = x

x = 3( y - 3 ) / ( 2y - 7 )

g⁻¹(x) = 3( x - 3 ) / ( 2x - 7 ), provided x ≠ 7/2

g⁻¹(-1) = 3( -1 - 3 ) / ( 2(-1) - 7 )

g⁻¹(-1) = 4/3

• its obvious. it spout 3,98.87645.54389.54. then it points west.

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And finally when compared to dantes s(&)^c =°✓€ it makes a plausible evauation of the suns accelleration of 2/π>¥.

pretty simple stuff.

• g: y=3+[x/(2x-3)]

=>

g^-1: x=3+[y/(2y-3)]

=>

(x-3)(2y-3)=y

=>

(2x-7)y=3x-9

=>

y=3(x-3)/(2x-7)

Thus,

g^-1(-1)=3(-1-3)/(-2-7)

=>

g^-1(-1)=4/3.

• It probably is not what you intended, but you have defined g as a quadratic function.

g(x) = 3+(x/2x-3)

g(x) = 3 + (x²/2 - 3)

g(x) = x²/2

I am only taking you at your word. If you meant something else, then please review the order of operations and try again.

Quadratic functions are not injective. Function g has no inverse.

• Or you can just set g(x) to -1 and soslve for x:

Presuming that is:

g(x) = 3 + x / (2x - 3)  <-- see the difference?  that's not what you wrote

We get:

-1 = 3 + x / (2x - 3)

-4 = x / (2x - 3)

-4(2x - 3) = x

-8x + 12 = x

12 = 9x

4/3 = x

I get the same thing as Astrid.

• Hint: As said previously, un-anon yourself and we will help you further.