Inverse of functions

Question

Find the inverse of f(x) = (x-2)^2 + 3 and f(x) = x-5 all under square root

Answer 1

f(x) = y = sqr(x-5) has an inverse where you switch x and y

which gives

x=sqr(y-5) square both sides (but realize you may be introducing an extraneous solution)

x² = y-5 add 5 to both sides

x² +5 = y

y= f(x)^-1 = x² +5

That at first, seems to be the answer, but the initial f(x) = only the positive square root of x-5, which is the top half of a parabola opening to the right. It's inverse is the mirror image reflected across the diagonal line through the origin. That gives only the right side of the upward opening parabola y=x² + 5

So while f(x)^-1 does = x²+5, that's only true for x>0 Forx<0, f(x)^-1 does not exist

At the step where you squared both sides, you introduced an extraneous solution which is not in the original problem.

Similarly for f(x)=y=(x-2)²+3 swtich x and y to find any potential inverse

and solve for y

x=(y-2)² + 3

(y-2)² = x-3

take square root of both sides to get, but realize squaring my introduce an extraneous solution

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

y = 2 + (x-3)^1/2 This looks like an answer y=f^-1=2 + (x-3)^1/2 is only an inverse function for x>3 and y>2

The y= 2 plus or minus is an inverse relation, but not inverse function. Eliminate the negative square root, and the positive square root is an inverse function for half of the original function for x>2

It helps to sketch the graph of everything, and put in a diagonal 45 degree line through the origin, and visualize where a mirror image would be below that line.

Answer 2

Function f(x) sqrt((x - 2)² + 3) has no inverse because it is not one-to-one function:

for any x ≠ 4 - x but f(x) = f(4 - x).

For second function f(x) = sqrt(x - 5); domain of f(x) is x ≥ 5 range is y ≥ 0;

To get inverse we interchange x and y and then solve for y: x = sqrt(y - 5); from here x² = y - 5, y = x² + 5.

Inverse function is f^-1(x) = x² + 5 with domain x ≥ 0.

Answer: 1st function has no inverse; inverse for second function f^-1(x) = x² + 5, x ≥ 0