The set of values of k, for which the function \(f(x)=kx^3+5\) is decreasing, is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \((-\infty, 0)\)
\(f'(x) = 3kx^2\)
Since \(x^2 \ge 0\), the sign of f′(x) is the sign of k (and f′(0) = 0 only at the single point x = 0).
For k < 0, \(f'(x) \le 0\) with equality only at x = 0, so f is decreasing on R. For k > 0 f is increasing, and for k = 0 f is constant.
So f is decreasing exactly when k < 0, that is, \(k \in (-\infty, 0)\).