Answer:
They are the same. If you were to graph them, they would overlap.
Step-by-step explanation:
Look at it this way:
[tex]\frac{3}{2} =1.5[/tex]
Insert this value:
[tex]f(x)=\frac{3}{2} x^2-4[/tex]
Now solve. Insert y:
[tex]y=\frac{3}{2}x^2-4[/tex]
Switch x and y:
[tex]x=\frac{3}{2}y^2-4[/tex]
Add 4 to both sides:
[tex]x+4=\frac{3}{2}y^2-4+4\\\\x+4=\frac{3}{2}y^2[/tex]
Use the rule [tex]\frac{a}{c} b=\frac{ab}{c}[/tex]:
[tex]x+4=\frac{3y^2}{2}[/tex]
Multiply both sides by 2:
[tex]2(x+4)=2(\frac{3y^2}{2})\\\\2x+8=3y^2[/tex]
Divide both sides by 3:
[tex]\frac{2x+8}{3} =\frac{3y^2}{3} \\\\\frac{2x+8}{3}=y^2[/tex]
Take the square root of both sides*:
[tex]\sqrt{\frac{2x+8}{3} } =\sqrt{y^2} \\\\y =\sqrt{\frac{2x+8}{3} } ,-\sqrt{\frac{2x+8}{3} }[/tex]
Therefore:
[tex]f^{-1}(x)=\sqrt{\frac{2x+8}{3} } ,-\sqrt{\frac{2x+8}{3} }[/tex]
This is not a function (by the way) because ±[tex]\sqrt{\frac{2x+8}{3} }[/tex] 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.
