If \(f(x) = mx^{2} - 6x - 3\) and \(f'(1) = 12\), find the value of the constant m.
Answer Details
The derivative of a quadratic function is obtained by differentiating the function term by term. Thus, differentiating the given function, we have:
$$f'(x) = 2mx - 6$$
We are also given that $f'(1) = 12$. Substituting $x=1$ and equating to 12, we have:
$$f'(1) = 2m(1) - 6 = 12$$
Simplifying this equation, we get:
$$2m = 18$$
Therefore, $m=9$.
Thus, the value of the constant $m$ is 9.