# What is the inverse of the function as defined by f(x)=square root(x^2+4), if x>0?

*print*Print*list*Cite

We have the function f(x) = sqrt(x^2 + 4) for values of x > 0

Let f(x) = y = sqrt(x^2 + 4)

Express x in terms of y

y = sqrt(x^2 + 4)

square both the sides

y^2 = (x^2 + 4)

=> x^2 = y^2 - 4

=> x = sqrt (y^2 - 4)

interchange x and y

=> y = sqrt (x^2 - 4)

**The inverse function of f(x) = sqrt(x^2 + 4) for values of x > 0 is f(x) = sqrt(x^2 - 4)**

f(x)= sqrt(x^2+4)

Let y= sqrt(X^2 + 4)

Now we need to isolate x on one side.

==> We will square both sides.

==> y^2 = x^2 + 4

Now we will subtract 4 from both sides.

==> y^2 - 4 = x^2

Now we will take theÂ square root.

==> x = sqrt(y^2 -4)

**Then the inverse function for f(x) is f^-1(x) = sqrt(x^2 -4)**

Let f(x) = y

y = sqrt(x^2+4)

We'll raise to square both sides to eliminate the radical:

y^2 = x^2 + 4

We'll use symmetrical property:

x^2 + 4 = y^2

We'll keep x^2 to the left side, shifting 1 to the right side:

x^2 = y^2 - 4

We'll take square both sides:

x = sqrt(y^2 - 4)

Since x > 0, we'll consider only the positive value x = sqrt(y^2 - 4).

The inverse of the given function is f^-1(x) = sqrt(x^2 - 4).