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hi please help im giving full points and brainliest:) Let y = 2(x – 3)2 – 6.

Part A: Is the given relation a function? Is it one-to-one? Explain completely. If it is not one-to-one, determine a possible restriction on the domain such that the relation is one-to-one. (5 points)

Part B: Determine y–1. Show all necessary calculations. (5 points)

Part C: Prove algebraically that y and y–1 are inverse functions. (5 points)

Respuesta :

Lenvy

Answer/Step-by-step explanation:

Part A:

Yes, it's a function. {Note: one x-value maps to only one y-value}

It's is not one-to-one. {one y-value sometimes can map to two x-values}

For the domain (4, + ∝) or for the domain ( - ∝,4), the relation is one-to-one

Part B:

[tex]y=2(x-4)^2-8[/tex]

[tex]y+8=2(x-4)^2[/tex]

[tex](x-4)^2=\frac{y}{2}+4[/tex]

[tex]x-4=[/tex] ± [tex]\sqrt{\frac{y}{2}+4} +4[/tex]

So [tex]y^-^1=[/tex] ± [tex]\sqrt{\frac{x}{2}=4} +4[/tex]

Part C: [tex]y(y^-^1)=2[/tex] ± [tex]\sqrt\frac{x}{2}+4x} +4-4)^2-8[/tex]

[tex]=2(\frac{x}{2}+4)-8[/tex]

[tex]=x+8-8=x[/tex]

So y and y^-^1 are inverse function

~Learn with Lenvy~

RevyBreeze

Answer/Step-by-step explanation:

Part A:

Yes, the given relation is a function

No, the given is not one-to-one.

The given relation is a function: True

The given one-to-one function: False

For the domain (3, + ∝) or the domain (- ∝,3)

the relation is one-to-one.

Part B:

[tex]y=2(x-3)^2-6[/tex]

[tex]y+6=2(x-3)[/tex]

[tex](x-2)^1=\frac{y}{2}+3[/tex]

[tex]x-2=[/tex] ±[tex]\sqrt{\frac{y}{2}+3}[/tex]

[tex]x=[/tex] ± [tex]\sqrt{\frac{y}{2}+2+2}[/tex]

So, [tex]y^-^1=[/tex] ± [tex]\sqrt{\frac{x}{2}+3}+3[/tex]

Part C:

[tex]y(y^-^1)=2([/tex]±[tex]\sqrt{\frac{x}{1}+3+3-3)^2-6}[/tex]

[tex]=2(\frac{x}{2}+3)-6[/tex]

[tex]x+6-6=x[/tex]

So y and y^-^1 are inverse function.

[RevyBreeze]

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