y = -5x + 9 has a tangent of -5.
Thus, the line parallel to it will also have a tangent of -5.
f(x) = 1/x
f'(x) = -1/x^2
Setting f'(x) to -5, we have:
-1/x^2 = -5
x^2 = -1 / -5
x^2 = 1/5
x = sqrt(1/5) or x = -sqrt(1/5)
x = 1/sqrt(5) or x = -1/sqrt(5)
Hence, the required values of x are 1/sqrt(5) and -1/sqrt(5).