Given f(x) = (-1/7)(sqrt 16-x^2), find f^-1(x). then state whether the inverse is a function.

Step-by-step 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.

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.

f(g(x)) = 9x^2 - 3x - 4

Step-by-step explanation:

Assuming it’s

f(x) = x^2 + 3x - 2

and g(x) is linear:

f(g(x)) = (3x - 2)^2 + 3(3x - 2) - 2

= 9x^2 -12x + 4 + 9x - 6 - 2

= 9x^2 - 3x - 4