Question

What’s the answer??(SOMEONE PLEASE HELP ME)

Answer 1

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

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

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

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

Now you need to isolate the "y"

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

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

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

$(x-8)^{3} = y -2$   Add 2 on both sides

$(x-8)^{3}+2 = y$

$f^{-1} = (x-8)^{3} + 2$

Answer 2

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.