Respuesta :

To find [tex]f^{-1}[/tex], you can switch the "x" and "f(x) or y" in the equation.

[tex]f(x) = \sqrt[3]{x-2} +8[/tex]

[tex]y = \sqrt[3]{x-2}+ 8[/tex]

[tex]x = \sqrt[3]{y-2}+8[/tex]

Now you need to isolate the "y"

[tex]x = \sqrt[3]{y-2}+8[/tex]   Subtract 8 on both sides

[tex]x - 8 = \sqrt[3]{y-2}[/tex]  Cube ( ³ ) each side to get rid of the ∛

[tex](x-8)^{3} = (\sqrt[3]{y-2}) ^{3}[/tex]

[tex](x-8)^{3} = y -2[/tex]   Add 2 on both sides

[tex](x-8)^{3}+2 = y[/tex]


[tex]f^{-1} = (x-8)^{3} + 2[/tex]


jcherry99

Remark

Interchange x and y

f(x) = y

y = ∛(x - 2) + 8      Now do the interchange

x = ∛(y - 2) + 8

(x - 8) = ∛(y - 2)     Cube both sides.

(x - 8)^3 = y - 2      Add 2 to both sides.

(x - 8)^3 + 2 = y = f-1(x)

x - 8 has a minus sign between the x and 8. B  is therefore wrong

There is no cube root in the inverse so C is incorrect

D is incorrect. The sign on the 2 is wrong.


The answer must therefore be A.

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