f(x)=3(x+x^{-1})

keep 3 alone for the calculation and consider it at last for final answer

⟹f^{'}(x)=1–x^{-2}

Critical point(s):

f^{′}(x)=0

1−x^{-}^{2}=0

⟹x=±1

f^{′′}(x)=2x^{-3}

f^{′′}(−1)=−2, f^{′′}(1)=2

Maximum value of f(x):f(−1)=−2 * 3 = -6

Minimum value of f(x):f(1)=2 * 3 = 6