Question 203520
One way to think of a function is that it takes an x-value and, some way or another, it comes up with a y-value. The inverse of that function does the inverse: it takes the y-values from the function and, some way or another, it comes up with the x-value the function used. In other words the roles of the x and y are swapped. And this is exactly how we can figure out an inverse:<ol><li>Write the function, using "y" instead of "x":
{{{y = -sqrt(x+2) + 1}}}</li><li>Swap the "x" and the "y". <b>This changes the equation</b> from the equation of the function to the equation of the inverse:
{{{x = -sqrt(y+2) + 1}}}</li><li>Solve the inverse equation for y:
Subtract -1 from both sides:
{{{x - 1 = -sqrt(y+2)}}}
Square both sides:
{{{(x - 1)^2 = y + 2}}}
{{{x^2 -2x + 1 = y + 2}}}
Subtract 2 from both sides:
{{{x^2 -2x -1 = y}}}</li><li>Replace the "y" with {{{f^(-1)(x)}}} (Note that Algebra.com's software incorrectly places a "*" between {{{f^(-1)}}} and {{{(x)}}}):
{{{x^2 -2x -1 = f^(-1)(x)}}}</li></ol>
The way to check is feed f(x) as an input to its inverse. It should work out to be x:
{{{f(x) = -sqrt(x+2) + 1}}}
{{{f^(-1)(x) = x^2 -2x -1}}}
{{{f^(-1)(f(x)) = (-sqrt(x+2) + 1)^2 - 2(-sqrt(x+2) + 1) - 1 = (-sqrt(x+2))^2 + 2(-sqrt(x+2))*1 + 1^2 - 2(-sqrt(x+2))  - 2 - 1 = (x+2) + -2sqrt(x+2) + 1 + 2sqrt(x+2) -2 -1 = x}}}
Check! (Note how everything except the "x" cancels at the end.)