Ta có : \(f\left(x\right)=2^{x-1}+2^{3-x}\ge2\sqrt{2^{x-1}.2^{3-x}}=4\)
Dấu bằng xảy ra khi và chỉ khi \(2^{x-1}=2^{3-x}\Leftrightarrow x-1=3-x\)
\(\Leftrightarrow x=2\)
Vậy Min \(f\left(x\right)=4\) khi x = 2
Ta có \(f'\left(x\right)=2^{x-1}\ln2-2^{3-x}\ln2=\left(2^{x-1}-2^{3-x}\right)\ln2=0\)
\(\Leftrightarrow2^{x-1}=^{3-x}\)
\(\Leftrightarrow x-1=3-x\)
\(\Leftrightarrow x=2\)
Mà \(\lim\limits_{x\rightarrow-\infty}f\left(x\right)=\lim\limits_{x\rightarrow-\infty}\left(2^{x-1}+2^{3-x}\right)=+\infty\)
\(\lim\limits_{x\rightarrow+\infty}f\left(x\right)=\lim\limits_{x\rightarrow+\infty}\left(2^{x-1}+2^{3-x}\right)=+\infty\)
Ta có bảng biến thiên :
Vậy Min f(x) = 4 khi x = 2
