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The local minimum value of f(x)=x3+243xf(x) = x^3 + \frac{243}{x} is

Solution

✅ Correct Option: 1

f′(x)=3x2−243x2=0⇒x4=81⇒x=±3f'(x)=3x^2-\frac{243}{x^2}=0\Rightarrow x^4=81\Rightarrow x=\pm3.

f′′(x)=6x+486x3f''(x)=6x+\frac{486}{x^3}, which is positive at x=3x=3, so x=3x=3 gives a local minimum.

f(3)=27+2433=27+81=108f(3)=27+\frac{243}{3}=27+81=108.

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