The...er...derivendicular

The information that the derivative is perpendicular to the function tells us that

$f'(x) = -\frac{1}{f''(x)}$

For simplicity substitute $y = f'(x)$ :

$y = -\frac{1}{y'}$ $y y' = -1$ $y \frac{dy}{dx} = -1$ $y dy = -dx$

Integrating:

$\frac 1 2 y^2 = -x + a$

$y = \sqrt{-2x + b}$

Where a and b are constants of integration. At this point, we can substitute back in for y:

$f'(x) = \sqrt{-2x + b}$

Integrating to find $f(x)$ :

$f(x) = - \frac 1 3 (-2x + b)^{\frac 3 2} + c$

where c is another constant of integration. Using the information that the domain of $f(x)$ is ${x \le 0}$ , we can infer that $b = 0$ . Using the information that $f(0) = 0$ , we can infer that $c = 0$ . This gives us the function

$f(x) = - \frac 1 3 (-2x)^{\frac 3 2}$

Evaluating this function at $x=-4.5$ , we get that

$f(-4.5) = -9$

The image shows a graph of this $f(x)$ and the corresponding $f'(x)$ .