For \(x \lt 0\), the local maximum value of \(f(x) = x + \displaystyle\frac{9}{x}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-6\)
Differentiating:
\( \displaystyle f'(x) = 1 - \frac{9}{x^2} \)
Setting f’(x) = 0 gives \(x^2 = 9\), so x = ±3. Since x < 0, the critical point is x = −3. The second derivative test:
\( \displaystyle f''(x) = \frac{18}{x^3} \)
\( \displaystyle f''(-3) = \frac{18}{-27} \lt 0 \)
So x = −3 is a point of local maximum, and the local maximum value is
\(f(-3) = -3 - 3 = -6\)
Why the other options are incorrect
Option (1) 6 — Incorrect. This is f(3), the local minimum value for x > 0, outside the given range.
Option (2) \(-3\) — Incorrect. This is the point x = −3, not the value of f there.
Option (4) 3 — Incorrect. This is the point x = 3, which is not negative and is not a value of f.