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3 years ago

Can someone help me answer this calculus I problem?

Mathematics

1 answer:

Elanso [62]3 years ago

6 0

Answer:

$f^{-1}(x) = \sqrt [7]{y - 3}$

Step-by-step explanation:

ƒ.(x) = x⁷ + 3

Rename f.(x) as y

y = x⁷ + 3

Solve for x :

x⁷ = y - 3

Take the seventh root of each side

$x = \sqrt [7]{y - 3}$

Switch x and y

$y = \sqrt [7]{x - 3}$

Rename y as the inverse function

$f^{-1}(x) = \sqrt [7]{y - 3}$

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$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$ (1)

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4 years ago

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