Đáp án:
Giải thích các bước giải:
1) `f(x)=4x+\frac{9}{x-1}`
TXĐ: `D=\mathbb{R} \\ {1}`
`f'(x)=4-\frac{9}{(x-1)^2}`
`f'(x)=0 ⇒` \(\left[ \begin{array}{l}x=\dfrac{5}{2}\ \in (1;+\infty)\\x=-\dfrac{1}{2}\ \notin (1;+\infty)\end{array} \right.\)
Ta có BBT:
\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$1$}&\text{}&\text{}\dfrac{5}{2}&\text{}\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}-&\text{$0$}&\text{}+&\\\hline \text{$y$}&\text{}+\infty&\text{}&\text{}&\text{}&\text{}+\infty\\&\text{}&\text{$\searrow$}&\text{}&\text{}\nearrow\\&\text{$$}&\text{}&\text{}16&\text{}&\text{}\\\hline \end{array}\)
Vậy `min_{(1;+\infty)} f(x)=16`
2)
`f(x)=x+4/x`
TXĐ: `D=\mathbb{R} \\ {0}`
`f'(x)=1-\frac{4}{x^2}`
`f'(x)=0 ⇒` \(\left[ \begin{array}{l}x=-2\ \in (-\infty;0)\\x=2\ \notin (-\infty;0)\end{array} \right.\)
Ta có BBT:
\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}-2&\text{}\text{}&\text{$0$}\\\hline \text{$y'$}&\text{}&\text{}+&\text{$0$}&\text{}-&\\\hline \text{$y$}&\text{}&\text{}&\text{}-4&\text{}&\text{}\\&\text{}&\text{$\nearrow$}&\text{}&\text{}\searrow\\&\text{$-\infty$}&\text{}&\text{}&\text{}&\text{}-\infty\\\hline \end{array}\)
Vậy `max_{(-\infty;0)} f(x)=-4`