What is the inverse function of f(x)=(x+1)^2f(x)=(x+1)2?

2 Answers
Sep 22, 2015

The function is not one-to-one. It does not have an inverse function. However,

Explanation:

If we restrict the domain to [-1,oo)[1,), then this new function has an inverse:

g(x)=(x+1)^2g(x)=(x+1)2 with x >= -1x1

y=(x+1)^2y=(x+1)2
if and only if
x+1 = sqrtyx+1=y (We do not need +-sqrty±y because of the restriction on xx.)

x=sqrty-1x=y1

g^-1(x) = -sqrtx-1g1(x)=x1

If we restrict the domain to (-oo,-1](,1], we get

h(x) = (x+1)^2h(x)=(x+1)2 with x <= -1x1

y=(x+1)^2y=(x+1)2
if and only if
x+1 = - sqrtyx+1=y (We need only -sqrtyy because of the restriction on xx.)

x= - sqrty-1x=y1

h^-1(x) = sqrtx-1h1(x)=x1

Sep 22, 2015

The function is not one-one and therefore there is not a unique inverse function.

Explanation:

First note that the slope (first derivative) is

f'(x) = 2(x + 1)(1) = 2x + 2f'(x)=2(x+1)(1)=2x+2 (chain rule).

Also note that

0 < 2 x + 2 0<2x+2

requires

-2 < 2x2<2x

that is

x > -1x>1

That is, the slope is negative for values of xx less than -11 and positive for values of xx greater than -11 (and has a stationary point at which the slope is zero at x = -1x=1).

That is, the function is not a strictly rising or a strictly descending one.

That is, it is not bijective (one-one).

That is, there is no single inverse function if the domain is taken as the reals.

The function is strictly descending on the open interval (-oo, -1)(,1) and strictly rising on the open interval (-1, oo)(1,) so there will be inverse functions on these domains.

Setting y = (x + 1)^2 y=(x+1)2

This implies

(x + 1) = +-sqrt(y)(x+1)=±y

which in turn implies

x = sqrt(y) - 1x=y1

or

x = -sqrt(y) - 1x=y1

Considering x = sqrt(y) - 1x=y1

Denoting the required inverse function by g(x)g(x), this may be rewritten as

g(x) = sqrt(x) - 1g(x)=x1

For g(x)g(x) to be a real function, the domain must be restricted to values of xx equal to or greater than 00.

Considering x = -sqrt(y) - 1x=y1

Denoting the required inverse function by h(x)h(x), this may be rewritten as

h(x) = - sqrt(x) - 1h(x)=x1

For h(x)h(x) to be a real function, the domain must be restricted to values of xx equal to or greater than 00.

Plotting both of these inverse functions on the same graph will yield a parabola "on its side", with apex at (0, -1)(0,1). That is, values of x > 0 x>0 will correspond to two values (one associated with g(x)g(x) and the other associated with h(x)h(x), reflecting the fact that f(x)f(x) is not a one-one function.