Find p(p^-1(x)) when p(x)=x^2-4
I tried solving this and I got x^4-x+20 but that has no solution so I'm a little confused.
for any function p, p(p^-1(x)) = p^-1(p(x)) = x
That's what inverse does!
p(x) = x^2-4
p^-1(x) = √(x+4)
so, p(p^-1) = (p^-1)^2-4 = (√(x+4))^2-4 = x+4-4 = x
p^-1(p) = √(p+4) = √(x^2-4+4) = √x^2 = x
now, you do have to watch out and beware of the fact that the domain of p^-1 is restricted.
Ohh okay got it thank you!
To find p(p^-1(x)), we need to first find the inverse function of p(x), denoted as p^-1(x).
To find the inverse function, we need to solve the equation y = p(x) for x. In this case, y = x^2 - 4.
Step 1: Set y = x^2 - 4.
Step 2: Interchange x and y.
y = x^2 - 4 switches to x = y^2 - 4.
Step 3: Solve for y.
Rearranging the equation, we have y^2 = x + 4.
Taking the square root of both sides, we get y = ±√(x + 4).
Since inverse functions are not multivalued, we choose one value for y. Let's choose y = √(x + 4).
Now that we have p^-1(x) = √(x + 4), we can substitute this into p(p^-1(x)) to find the expression:
p(p^-1(x)) = p(√(x + 4)).
Substituting p(x) = x^2 - 4, we have:
p(p^-1(x)) = (√(x + 4))^2 - 4.
Simplifying the expression:
p(p^-1(x)) = x + 4 - 4.
p(p^-1(x)) = x.
So, the final answer is p(p^-1(x)) = x.