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Someone check my work, please :)

Find the inverse:

f(x)=1.5x^2-4

My work:

y=1.5x^2-4\\\\x=1.5y^2-4\\\\x+4=1.5y^2\\\\(x+4)/(1.5) =y^2\\\\y=\sqrt{(x+4)/(1.5) } \\\\f^(-1)(x)=y=\sqrt{(x+4)/(1.5) }

However, the answer sheet says the answer is
f^(-1)(x)=\sqrt{(2x+8)/(3) }. I did notice that if you multiply each number by 2 you get the numbers given in the answers:

1*2=2\\\\4*2=8\\\\1.5*2=3

Thanks :)

1 Answer

7 votes

Answer:

They are the same. If you were to graph them, they would overlap.

Explanation:

Look at it this way:


(3)/(2) =1.5

Insert this value:


f(x)=(3)/(2) x^2-4

Now solve. Insert y:


y=(3)/(2)x^2-4

Switch x and y:


x=(3)/(2)y^2-4

Add 4 to both sides:


x+4=(3)/(2)y^2-4+4\\\\x+4=(3)/(2)y^2

Use the rule
(a)/(c) b=(ab)/(c):


x+4=(3y^2)/(2)

Multiply both sides by 2:


2(x+4)=2((3y^2)/(2))\\\\2x+8=3y^2

Divide both sides by 3:


(2x+8)/(3) =(3y^2)/(3) \\\\(2x+8)/(3)=y^2

Take the square root of both sides*:


\sqrt{(2x+8)/(3) } =√(y^2) \\\\y =\sqrt{(2x+8)/(3) } ,-\sqrt{(2x+8)/(3) }

Therefore:


f^(-1)(x)=\sqrt{(2x+8)/(3) } ,-\sqrt{(2x+8)/(3) }

This is not a function (by the way) because ±
\sqrt{(2x+8)/(3) } means that there are two y values per x value. If you were to graph this, it would fail the vertical line test, which proves that it is not a function.

:Done

*You need to have the ± symbol when you take a square root because, although 2×2=4, (-2)×(-2)=4 as well. The square root of 4 could be either -2 or positive 2. The same can be said for the above result.

User Mirko Catalano
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