Minimum value of \( f(x)=x^{2}+\displaystyle\frac{250}{x} \), \( x\ne 0 \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 75
Here x may also be negative, and as \( x\to 0^{-} \) the term \( \frac{250}{x} \) becomes very large and negative. So f has no least value over its whole domain, and the minimum value meant is the local minimum value, as usual in such questions.
\( \displaystyle f'(x)=2x-\frac{250}{x^{2}} \)
Setting \( f'(x)=0 \):
\( \displaystyle 2x^{3}=250 \)
\( \displaystyle x^{3}=125\ \Rightarrow\ x=5 \)
\( \displaystyle f''(x)=2+\frac{500}{x^{3}} \)
\( \displaystyle f''(5)=2+4=6\gt 0 \)
So f has a local minimum at x = 5, and the minimum value is
\( \displaystyle f(5)=25+\frac{250}{5}=75 \)
Shortcut: For \( x\gt 0 \), apply AM-GM to \( x^{2},\ \frac{125}{x},\ \frac{125}{x} \):
\( \displaystyle f(x)\ge 3\sqrt[3]{x^{2}\cdot\frac{125}{x}\cdot\frac{125}{x}}=3\sqrt[3]{125^{2}}=75 \)