What is the inverse of y=(5x^3-2)/(x^3+1) ?

The inverse of y=(5x^3-2)/(x^3+1) is cube root of (-2-x)/(x-5)

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inverse of y=(5x^3-2)/(x^3+1)

inverse

y = \frac{5 x ^{3} - 2}{x ^{3} + 1}

Solution steps

y = \frac{5 x ^{3} - 2}{x ^{3} + 1}

Replace

x

with

y

x = \frac{5 y ^{3} - 2}{y ^{3} + 1}

Solve for

y, x = \frac{5 y ^{3} - 2}{y ^{3} + 1}

f (x) = - x^{2} - 3 x + 4

100 - x^{2}

- \frac{1}{3} (x - 2)^{2} - 4

f (x) = 2 x + 4 - 1

f (x) = (x + 3) 4

What is the inverse of y=(5x^3-2)/(x^3+1) ?
The inverse of y=(5x^3-2)/(x^3+1) is cube root of (-2-x)/(x-5)