How do you verify that $f (x) = 3 x^{4} + 1, x \geq 0; g (x) = \sqrt[4]{\frac{1}{3} x - \frac{1}{3}}$ $f(x)=3x^4+1, x>=0; g(x)=root4(1/3x-1/3)$ are inverses?

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How do you verify that $f (x) = 3 x^{4} + 1, x \geq 0; g (x) = \sqrt[4]{\frac{1}{3} x - \frac{1}{3}}$ $f(x)=3x^4+1, x>=0; g(x)=root4(1/3x-1/3)$ are inverses?

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Answer 1 · 2017-08-13T19:32:29

If $f (g (x)) = x$ $f(g(x)) = x$ , then they are indeed inverses.

$f (g (x)) = 3 {(\sqrt[4]{\frac{1}{3} x - \frac{1}{3}})}^{4} + 1$ $f(g(x)) = 3(root(4)(1/3x- 1/3))^4 + 1$

$f (g (x)) = 3 {({(\frac{1}{3} x - \frac{1}{3})}^{\frac{1}{4}})}^{4} + 1$ $f(g(x)) = 3((1/3x - 1/3)^(1/4))^4 + 1$

$f (g (x)) = 3 (\frac{1}{3} x - \frac{1}{3}) + 1$ $f(g(x)) = 3(1/3x - 1/3) + 1$

$f (g (x)) = x - 1 + 1$ $f(g(x)) = x - 1 + 1$

$f (g (x)) = x$ $f(g(x)) = x$

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How do you verify that $f (x) = 3 x^{4} + 1, x \geq 0; g (x) = \sqrt[4]{\frac{1}{3} x - \frac{1}{3}}$ $f(x)=3x^4+1, x>=0; g(x)=root4(1/3x-1/3)$ are inverses?

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How do you verify that $f (x) = 3 x^{4} + 1, x \geq 0; g (x) = \sqrt[4]{\frac{1}{3} x - \frac{1}{3}}$ $f(x)=3x^4+1, x>=0; g(x)=root4(1/3x-1/3)$ are inverses?