+128707 Geno's answer is correct, but let me add one proviso.
We must restrict the domain on the inverse to x ≥ - 1.
The reason for this is that if (a,b) is on the inverse graph, then (b, a ) must be on the original graph. And the minimum point on y = x^(3/2) - 1 = (0,-1). Notice that if we put -9 into y = (x + 1)^(2/3), the result is 4. So the point (-9,4) is on the inverse graph, but the point (4, -9) isn't on the original graph. And notice that if we put -1 into the inverse, it returns 0. And for all x values greaer than -1 on the inverse, the two graphs will maintain this (a,b), (b,a) relationship.
+23247 f(x) = x^(3/2) - 1
y = x^(3/2) - 1
Step #1: interchange x and y:
x = y^(3/2) -1
Step #2: solve for y:
x+ 1 = y^(3/2)
Square both sides: (x + 1)^2 = y^3
Find the cube root of both sides: (x + 1) ^(2/3) = y
Answer: y = (x + 1) ^(2/3)
If you have any questions, please post back.
+128707 Geno's answer is correct, but let me add one proviso.
We must restrict the domain on the inverse to x ≥ - 1.
The reason for this is that if (a,b) is on the inverse graph, then (b, a ) must be on the original graph. And the minimum point on y = x^(3/2) - 1 = (0,-1). Notice that if we put -9 into y = (x + 1)^(2/3), the result is 4. So the point (-9,4) is on the inverse graph, but the point (4, -9) isn't on the original graph. And notice that if we put -1 into the inverse, it returns 0. And for all x values greaer than -1 on the inverse, the two graphs will maintain this (a,b), (b,a) relationship.
