Answer:
- f^-1(x) = ±√(16 -49x^2) . . . . -4/7 ≤ x ≤ 0
- not a function
Explanation:
The usual method of finding the inverse of a function is to solve for y the equation ...
x = f(y)
The inverse is only defined on the range of the original function, so for ...
0 ≥ x ≥ -4/7
Here, that looks like ...
x = (-1/7)√(16 -y^2)
-7x = √(16 -y^2) . . . multiply by -7
49x^2 = 16 -y^2 . . . square both sides
y^2 + 49x^2 = 16 . . . add y^2
y^2 = 16 -49x^2 . . . . subtract 49x^2
y = ±√(16 -49x^2) . . . . take the square root
So, the inverse function is ...
f^-1(x) = ±√(16 -49x^2) . . . . . defined only for -4/7 ≤ x ≤ 0
This gives two output values for each input value, so is not a function.
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The original f(x) is the bottom half of an ellipse, so does not pass the horizontal line test. It cannot have an inverse function.