What is the inverse of (x^2(x+1))/(x+1) ?

The inverse of (x^2(x+1))/(x+1) is sqrt(x),-sqrt(x)

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inverse of (x^2(x+1))/(x+1)

inverse

\frac{x ^{2} ( x + 1 )}{x + 1}

x, - x

Solution steps

y = \frac{x ^{2} ( x + 1 )}{x + 1}

Replace

x

with

y

x = \frac{y ^{2} ( y + 1 )}{y + 1}

Solve for

y, x = \frac{y ^{2} ( y + 1 )}{y + 1}

x, - x

x^{3} - 5 x

\frac{2 ( x + 2 )}{x ^{2}}

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

f (x) = - sin (4 x)

\frac{2 x + 3}{4}

What is the inverse of (x^2(x+1))/(x+1) ?
The inverse of (x^2(x+1))/(x+1) is sqrt(x),-sqrt(x)