2 Answers

Kakadu · Answer 1 · 2019-09-04T01:29:59+0000

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

Geobio Boo · Answer 2 · 2019-09-09T06:25:58+0000

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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