Cho f(x)= 2/3x2 -1/5x.Tìm x để f(x) đạt GTNN và tính GTNN đó?
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
2 tháng 11 2019
\(f\left(x\right)=\left|x-2015\right|+\left|x+2016\right|\)
a) Ta có: \(\left|x\right|=\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
+) Với \(x=\frac{1}{2}\):
\(f\left(\frac{1}{2}\right)=\left|\frac{1}{2}-2015\right|+\left|\frac{1}{2}+2016\right|=2\)
+) Với \(x=-\frac{1}{2}\)
\(f\left(-\frac{1}{2}\right)=\left|-\frac{1}{2}-2015\right|+\left|-\frac{1}{2}+2016\right|=0\)
KN
2 tháng 11 2019
c) Áp dụng BĐT |x| + |y| \(\ge\)|x + y|, ta được:
\(f\left(x\right)=\left|x-2015\right|+\left|x+2016\right|=\left|2015-x\right|+\left|x+2016\right|\)
\(\ge\left|\left(2015-x\right)+\left(x+2016\right)\right|=\left|4031\right|=4031\)
(Dấu "="\(\Leftrightarrow\left(2015-x\right)\left(x+2016\right)\ge0\)
TH1: \(\hept{\begin{cases}2015-x\ge0\\x+2016\ge0\end{cases}}\Leftrightarrow-2016\le x\le2015\)
TH2: \(\hept{\begin{cases}2015-x\le0\\x+2016\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2015\\x\le-2016\end{cases}}\left(L\right)\))
Vậy \(f\left(x\right)_{min}=4031\Leftrightarrow-2016\le x\le2015\)
TN
9 tháng 8 2016
f(x)= 2x2-7x+1
\(=2\left(x^2-\frac{7x}{2}+\frac{1}{2}\right)\)
\(=2\left(x^2-\frac{7x}{2}+\frac{49}{16}\right)-\frac{41}{8}\)
\(=2\left(x-\frac{7}{4}\right)^2-\frac{41}{8}\ge0-\frac{41}{8}=-\frac{41}{8}\)
Dấu = khi \(2\left(x-\frac{7}{4}\right)^2=0\Leftrightarrow x-\frac{7}{4}\Leftrightarrow x=\frac{7}{4}\)
Vậy...
TN
25 tháng 7 2017
1,\(f\left(x\right)=3x^2-2x-7\)
\(=3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{22}{3}\)
\(=2\left(x-\dfrac{1}{3}\right)^2-\dfrac{22}{3}\ge-\dfrac{22}{3}\forall x\)
Vậy GTNN của biểu thức là \(-\dfrac{22}{3}\) khi \(x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)
\(b,f\left(x\right)=5x^2+7x=5\left(x^2+\dfrac{7}{5}x+\dfrac{49}{100}\right)-\dfrac{49}{20}\)\(=5\left(x+\dfrac{7}{10}\right)^2-\dfrac{49}{20}\ge-\dfrac{49}{20}\forall x\)
Vậy Giá trị nhỏ nhất của biểu thức là \(-\dfrac{49}{20}\) khi \(x+\dfrac{7}{10}=0\Rightarrow x=-\dfrac{7}{10}\)
\(c,f\left(x\right)=-5x^2+9x-2=-5\left(x^2-\dfrac{9}{5}x+\dfrac{81}{100}\right)+\dfrac{41}{20}\)\(=-5\left(x-\dfrac{9}{10}\right)^2+\dfrac{41}{20}\le\dfrac{41}{20}\forall x\)
Vậy GTLN của biểu thức là \(\dfrac{41}{20}\) khi \(x-\dfrac{9}{10}=0\Rightarrow x=\dfrac{9}{10}\)
\(d,f\left(x\right)=-7x^2+3x=-7\left(x^2-\dfrac{3}{7}x+\dfrac{9}{196}\right)+\dfrac{9}{28}\)\(=-7\left(x-\dfrac{3}{14}\right)^2+\dfrac{9}{28}\le\dfrac{9}{28}\forall x\)
Vậy GTLN của biểu thức là \(\dfrac{9}{28}\) khi \(x-\dfrac{3}{14}=0\Rightarrow x=\dfrac{3}{14}\)
25 tháng 7 2017
1/ \(f\left(x\right)=3x^2-2x-7\)
\(=3\left(x^2-\dfrac{2}{3}x-7\right)\)
\(=3\left(x^2-\dfrac{2}{3}+\dfrac{1}{9}-\dfrac{64}{9}\right)\)
\(=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{64}{3}\)
Ta có: \(3\left(x-\dfrac{1}{3}\right)^2\ge0\forall x\Rightarrow3\left(x-\dfrac{1}{3}\right)^2-\dfrac{64}{3}\ge-\dfrac{64}{3}\forall x\)
Dấu "=" xảy ra khi \(x-\dfrac{1}{3}=0\) hay \(x=\dfrac{1}{3}\)
Vậy MINf(x) = \(-\dfrac{64}{3}\) khi x = \(\dfrac{1}{3}\).
2/ \(f\left(x\right)=5x^2+7x\)
\(=5\left(x^2+\dfrac{7}{5}x\right)=5\left(x^2+\dfrac{7}{5}x+\dfrac{49}{100}-\dfrac{49}{100}\right)\)
\(=5\left(x+\dfrac{7}{10}\right)^2-\dfrac{49}{20}\)
Ta có: \(5\left(x+\dfrac{7}{10}\right)^2\ge0\forall x\Rightarrow5\left(x+\dfrac{7}{10}\right)^2-\dfrac{49}{20}\ge-\dfrac{49}{20}\forall x\)
Dấu "=" xảy ra khi \(x+\dfrac{7}{10}=0\) hay \(x=-\dfrac{7}{10}\)
Vậy MINf(x) = \(-\dfrac{49}{20}\) khi x = \(-\dfrac{7}{10}\).
1/ \(f\left(x\right)=-5x^2+9x-2\)
\(=-5\left(x^2-\dfrac{9}{5}x+\dfrac{2}{5}\right)\)
\(=-5\left(x^2-\dfrac{9}{5}x+\dfrac{81}{100}-\dfrac{41}{100}\right)\)
\(=-5\left(x-\dfrac{9}{10}\right)^2+\dfrac{41}{20}\)
Ta có: \(-5\left(x-\dfrac{9}{10}\right)^2\le0\forall x\Rightarrow-5\left(x-\dfrac{9}{10}\right)^2+\dfrac{41}{20}\le\dfrac{41}{20}\forall x\)
Dấu "=" xảy ra khi \(x-\dfrac{9}{10}=0\) hay \(x=\dfrac{9}{10}\)
Vậy MAXf(x) = \(\dfrac{41}{20}\) khi x = \(\dfrac{9}{10}\)
2/ \(f\left(x\right)=-7x^2+3x=-7\left(x^2-\dfrac{3}{7}x+\dfrac{9}{196}\right)+\dfrac{9}{28}\)
\(=-7\left(x-\dfrac{3}{14}\right)^2+\dfrac{9}{28}\)
Ta có: \(-7\left(x-\dfrac{3}{14}\right)^2\le0\forall x\Rightarrow-7\left(x-\dfrac{3}{14}\right)^2+\dfrac{9}{28}\le\dfrac{9}{28}\forall x\)
Dấu "=" xảy ra khi \(x-\dfrac{3}{14}=0\) hay x = \(\dfrac{3}{14}\)
Vậy MAXf(x) = \(\dfrac{9}{28}\) khi x = \(\dfrac{3}{14}\).
11 tháng 7 2017
1. Ta có: \(f\left(x\right)=9x^2-12x+1=\left(3x\right)^2-2.3x.2+2^2-3\)
\(=\left(3x-2\right)^2-3\)
Vì \(\left(3x-2\right)^2\ge0\) với mọi x \(\Rightarrow\left(3x-2\right)^2-3\ge-3\) hay \(f\left(x\right)\ge-3\)
Dấu ''='' xảy ra \(\Leftrightarrow\left(3x-2\right)^2=0\Rightarrow3x-2=0\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)
Vậy min f(x) =-3 khi \(x=\dfrac{2}{3}\)
2. Ta có: \(f\left(x\right)=2x^2-7x+5=2.\left(x^2-3,5x\right)+5=2.\left(x^2-2.x.1,75+1,75^2\right)-2.1,75^2+5\)
\(=2.\left(x-1,75\right)^2-1,125\)
Vì \(2.\left(x-1,75\right)^2\ge0\Rightarrow2.\left(x-1,75\right)^2-1,125\ge-1,125\Rightarrow f\left(x\right)\ge-1,125\)
Dấu ''='' xảy ra \(\Leftrightarrow2.\left(x-1,75\right)^2=0\Rightarrow x-1,75=0\Rightarrow x=1,75\)
Vậy min f(x)=-1,125 khi x=1,75
3.\(3x^2-10x=3.\left(x^2-\dfrac{10}{3}x\right)=3.\left(x^2-2.x.\dfrac{5}{3}\right)\)
\(=3.\left[x^2-2.x.\dfrac{5}{3}+\left(\dfrac{5}{3}\right)^2\right]-3.\left(\dfrac{5}{3}\right)^2\)
\(=3.\left(x-\dfrac{5}{3}\right)^2-\dfrac{25}{3}\)
Vì \(3.\left(x-\dfrac{5}{3}\right)^2\ge0\Rightarrow3.\left(x-\dfrac{5}{3}\right)^2-\dfrac{25}{3}\ge-\dfrac{25}{3}\Rightarrow f\left(x\right)\ge-\dfrac{25}{3}\)
Dấu ''='' xảy ra \(\Leftrightarrow3.\left(x-\dfrac{5}{3}\right)^2=0\Rightarrow x-\dfrac{5}{3}=0\Rightarrow x=\dfrac{5}{3}\)
Vậy min f(x)=\(-\dfrac{25}{3}\) khi \(x=\dfrac{5}{3}\)
