Find (f^-1)(−2) if f(x) = x.sqrt(3 + x^2) and if f ^-1(x) exists.

Answer 1

I will show an easy way (guess at a solution of f(x) = -2, show that it is unique, and thus have found f^-1(-2))

Note that f(x) is a monotone increasing function of x, so if we find an x such that f(x) = -2, then that value of x will be the inverse.

To show that f(x) is monotone increasing, we could either make statements about f(x) itself or note that, for f(x) = x sqrt(3+x^2), f'(x) = sqrt(3+x^2) + x^2/sqrt(3+x^2) =

(3+2x^2)/(3+x^2)^(1/2) exists and is positive everywhere.

Note that f(0) = 0. Thus, x and f(x) have the same sign. This will be useful later.

Now, let x = -1. Then f(x) = -1 * sqrt(3+(-1)^2) = -1 * sqrt(3+1) = -1 * sqrt(4) = -1 * 2 = -2.

Thus, f^-1(-2) = -1

A harder way: y= x √(3+x^2)square both sides. Then, y^2 = x^2(3+x^2) = 3x^2 + x^4.

x^4 + 3 x^2 - y^2 = 0

Let z = x^2.

Then, z^2 + 3z - y^2 = 0

z = -3/2 ± √(9/4 + y^2)

As z = x^2, we only take the positive root. Thus, z = -3/2 + √(9/4 + y^2)

As x^2 = z, x = √(-3/2 + √(9/4 + y^2)) We now figure out the unique square root (squaring often introduces extra values that are not solutions and need to be eliminated).

As mentioned above, x and y have the same sign.

Thus, x = signum(y)* √(-3/2 + √(9/4 + y^2))

Then, f^-1(x) = signum(x)* √(-3/2 + √(9/4 + x^2)).

Thus, f^-1(-2) = -1 √(-3/2 + √(9/4 + x^2)) = -1 √(-3/2 + √(9/4 + 4)) = -1(√(-3/2 + √25/4) = -1 (√(-3/2 + 5/2) =

-1 sqrt(1) = -1