Toán tìm min hoặc max của hàm số y =(X+2)\(X bình -X +3) 01/07/2021 By Natalia tìm min hoặc max của hàm số y =(X+2)\(X bình -X +3)
Đáp án: $Miny = \dfrac{{ – 1}}{{11}} \Leftrightarrow x = – 5$ $Maxy = 1 \Leftrightarrow x = 1$ Giải thích các bước giải: Ta có: $A = \dfrac{{x + 2}}{{{x^2} – x + 3}}$ +) Tìm Max: $\begin{array}{l}A – 1 = \dfrac{{x + 2}}{{{x^2} – x + 3}} – 1\\ = \dfrac{{x + 2 – \left( {{x^2} – x + 3} \right)}}{{{x^2} – x + 3}}\\ = \dfrac{{ – {x^2} + 2x – 1}}{{{x^2} – x + 3}}\\ = \dfrac{{ – {{\left( {x – 1} \right)}^2}}}{{{x^2} – x + 3}}\end{array}$ Nhận xét: $\begin{array}{l}\left\{ \begin{array}{l}{\left( {x – 1} \right)^2} \ge 0,\forall x\\{x^2} – x + 3 = {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{4} > 0,\forall x\end{array} \right.\\ \Rightarrow – {\left( {x – 1} \right)^2} \le 0,\forall x\\ \Rightarrow \dfrac{{ – {{\left( {x – 1} \right)}^2}}}{{{x^2} – x + 3}} \le 0,\forall x\\ \Rightarrow y – 1 \le 0\\ \Rightarrow y \le 1\\ \Rightarrow Maxy = 1\end{array}$ Dấu bằng xảy ra $\begin{array}{l} \Leftrightarrow {\left( {x – 1} \right)^2} = 0\\ \Leftrightarrow x = 1\end{array}$ Vậy $Maxy = 1 \Leftrightarrow x = 1$ +) Tìm Min: $\begin{array}{l}y + \dfrac{1}{{11}} = \dfrac{{x + 2}}{{{x^2} – x + 3}} + \dfrac{1}{{11}}\\ = \dfrac{{{x^2} – x + 3 + 11\left( {x + 2} \right)}}{{11\left( {{x^2} – x + 3} \right)}}\\ = \dfrac{{{x^2} + 10x + 25}}{{11\left( {{x^2} – x + 3} \right)}}\\ = \dfrac{{{{\left( {x + 5} \right)}^2}}}{{11\left( {{x^2} – x + 3} \right)}}\end{array}$ Nhận xét: $\begin{array}{l}\left\{ \begin{array}{l}{\left( {x + 5} \right)^2} \ge 0,\forall x\\{x^2} – x + 3 = {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{4} > 0,\forall x\end{array} \right.\\ \Rightarrow \dfrac{{{{\left( {x + 5} \right)}^2}}}{{11\left( {{x^2} – x + 3} \right)}} \ge 0\\ \Rightarrow y + \dfrac{1}{{11}} \ge 0\\ \Rightarrow Miny = \dfrac{{ – 1}}{{11}} \Leftrightarrow x = – 5\end{array}$ $\begin{array}{l} \Leftrightarrow {\left( {x + 5} \right)^2} = 0\\ \Leftrightarrow x = – 5\end{array}$ Vậy $Miny = \dfrac{{ – 1}}{{11}} \Leftrightarrow x = – 5$ Trả lời
Đáp án:
$Miny = \dfrac{{ – 1}}{{11}} \Leftrightarrow x = – 5$
$Maxy = 1 \Leftrightarrow x = 1$
Giải thích các bước giải:
Ta có:
$A = \dfrac{{x + 2}}{{{x^2} – x + 3}}$
+) Tìm Max:
$\begin{array}{l}
A – 1 = \dfrac{{x + 2}}{{{x^2} – x + 3}} – 1\\
= \dfrac{{x + 2 – \left( {{x^2} – x + 3} \right)}}{{{x^2} – x + 3}}\\
= \dfrac{{ – {x^2} + 2x – 1}}{{{x^2} – x + 3}}\\
= \dfrac{{ – {{\left( {x – 1} \right)}^2}}}{{{x^2} – x + 3}}
\end{array}$
Nhận xét:
$\begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x – 1} \right)^2} \ge 0,\forall x\\
{x^2} – x + 3 = {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{4} > 0,\forall x
\end{array} \right.\\
\Rightarrow – {\left( {x – 1} \right)^2} \le 0,\forall x\\
\Rightarrow \dfrac{{ – {{\left( {x – 1} \right)}^2}}}{{{x^2} – x + 3}} \le 0,\forall x\\
\Rightarrow y – 1 \le 0\\
\Rightarrow y \le 1\\
\Rightarrow Maxy = 1
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow {\left( {x – 1} \right)^2} = 0\\
\Leftrightarrow x = 1
\end{array}$
Vậy $Maxy = 1 \Leftrightarrow x = 1$
+) Tìm Min:
$\begin{array}{l}
y + \dfrac{1}{{11}} = \dfrac{{x + 2}}{{{x^2} – x + 3}} + \dfrac{1}{{11}}\\
= \dfrac{{{x^2} – x + 3 + 11\left( {x + 2} \right)}}{{11\left( {{x^2} – x + 3} \right)}}\\
= \dfrac{{{x^2} + 10x + 25}}{{11\left( {{x^2} – x + 3} \right)}}\\
= \dfrac{{{{\left( {x + 5} \right)}^2}}}{{11\left( {{x^2} – x + 3} \right)}}
\end{array}$
Nhận xét:
$\begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x + 5} \right)^2} \ge 0,\forall x\\
{x^2} – x + 3 = {\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{4} > 0,\forall x
\end{array} \right.\\
\Rightarrow \dfrac{{{{\left( {x + 5} \right)}^2}}}{{11\left( {{x^2} – x + 3} \right)}} \ge 0\\
\Rightarrow y + \dfrac{1}{{11}} \ge 0\\
\Rightarrow Miny = \dfrac{{ – 1}}{{11}} \Leftrightarrow x = – 5
\end{array}$
$\begin{array}{l}
\Leftrightarrow {\left( {x + 5} \right)^2} = 0\\
\Leftrightarrow x = – 5
\end{array}$
Vậy $Miny = \dfrac{{ – 1}}{{11}} \Leftrightarrow x = – 5$