Let \(f(x) = x^2+\displaystyle\frac{250}{x}\) be any function defined on \(R - \{0\}\), where \(R\) is the set of real numbers. Then which of the following are TRUE?
(A) \(f'(x) = 2x +\displaystyle\frac{250}{x^2}\)
(B) \(x = 5\) is the only critical point of f(x)
(C) minimum value of f(x) is 75
(D) maximum value of f(x) is 50
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B) and (C) only
\( \displaystyle f'(x) = 2x - \frac{250}{x^2} \)
Putting \(f'(x) = 0\):
\(2x^3 = 250\)
\(x^3 = 125\)
\(x = 5\)
This is the only real root, so \(x = 5\) is the only critical point.
\( \displaystyle f''(x) = 2 + \frac{500}{x^3} \)
\(f''(5) = 2 + 4 = 6 \gt 0\)
So f has a local minimum at \(x = 5\):
\(f(5) = 25 + 50 = 75\)
With no other critical point, f has no local maximum.
As x approaches 0 from the left, \(\frac{250}{x}\) becomes very large and negative, so f has no least value on the whole of \(R - \{0\}\). The minimum value in statement (C) is therefore the local minimum value, as is usual in such questions.
So \(f'(x) = 2x - \frac{250}{x^2}\), the only critical point is \(x = 5\), the local minimum value is 75, and f has no maximum value.
Checking the statements
(A) \(f'(x) = 2x +\frac{250}{x^2}\) — Incorrect. The derivative of \(\frac{250}{x}\) is \(-\frac{250}{x^2}\).
(B) \(x = 5\) is the only critical point of f(x) — Correct. \(x^3 = 125\) has the single real root 5.
(C) minimum value of f(x) is 75 — Correct. The local minimum value is \(f(5) = 75\), with \(f''(5) = 6 \gt 0\).
(D) maximum value of f(x) is 50 — Incorrect. The only critical point gives a minimum, so f has no maximum value.