cho hàm số f(t)= 2t – $\frac{3}{8}$$t^{2}$ , t>0 tìm gtln của f(t) 27/11/2021 Bởi aikhanh cho hàm số f(t)= 2t – $\frac{3}{8}$$t^{2}$ , t>0 tìm gtln của f(t)
$f(t)=\dfrac{-3}{8}t^2+2t$ $-f(t)=\dfrac{3}{8}t^2-2t$ $=\Big(\dfrac{\sqrt6}{4}t\Big)^2-2.\dfrac{\sqrt6}{4}.\dfrac{2\sqrt6}{3}+\dfrac{8}{3}-\dfrac{8}{3}$ $=\Big(\dfrac{\sqrt6}{4}t-\dfrac{2\sqrt6}{3}\Big)^2-\dfrac{8}{3}\ge \dfrac{-8}{3}$ $\Leftrightarrow f(t)=-\Big(\dfrac{\sqrt6}{4}t-\dfrac{2\sqrt6}{3}\Big)^2+\dfrac{8}{3}\le \dfrac{8}{3}$ $\max =\dfrac{8}{3}\Leftrightarrow t=\dfrac{8}{3}$ (TM) Bình luận
Đáp án: Giải thích các bước giải: ta có f( t) = 2t – $\frac{3}{8}$ t² = – ( $\frac{3}{8}$ t² -2t) có: $\frac{3}{8}$ t² -2t = $\frac{3}{8}$ t² – 2t× $\sqrt[]{}$ $\frac{3}{8}$ ×$\sqrt[]{}$ $\frac{8}{3}$ + $\frac{8}{3}$ -$\frac{8}{3}$ =( $\sqrt[]{}$ $\frac{3}{8}$ t-$\sqrt[]{}$ $\frac{8}{3}$ )² – $\frac{8}{3}$ ta có ($\sqrt[]{}$ $\frac{3}{8}$t -$\sqrt[]{}$ $\frac{8}{3}$ )² ≥0 ⇒ ($\sqrt[]{}$ $\frac{3}{8}$ t-$\sqrt[]{}$ $\frac{8}{3}$ )² -$\frac{8}{3}$ $\geq$ $\frac{-8}{3}$ ⇒ – ( $\sqrt[]{}$ $\frac{3}{8}$t -$\sqrt[]{}$ $\frac{8}{3}$ )² -$\frac{8}{3}$ $\leq$ ) ≤ $\frac{8}{3}$ ⇒max = $\frac{8}{3}$ Bình luận
$f(t)=\dfrac{-3}{8}t^2+2t$
$-f(t)=\dfrac{3}{8}t^2-2t$
$=\Big(\dfrac{\sqrt6}{4}t\Big)^2-2.\dfrac{\sqrt6}{4}.\dfrac{2\sqrt6}{3}+\dfrac{8}{3}-\dfrac{8}{3}$
$=\Big(\dfrac{\sqrt6}{4}t-\dfrac{2\sqrt6}{3}\Big)^2-\dfrac{8}{3}\ge \dfrac{-8}{3}$
$\Leftrightarrow f(t)=-\Big(\dfrac{\sqrt6}{4}t-\dfrac{2\sqrt6}{3}\Big)^2+\dfrac{8}{3}\le \dfrac{8}{3}$
$\max =\dfrac{8}{3}\Leftrightarrow t=\dfrac{8}{3}$ (TM)
Đáp án:
Giải thích các bước giải:
ta có f( t) = 2t – $\frac{3}{8}$ t² = – ( $\frac{3}{8}$ t² -2t)
có: $\frac{3}{8}$ t² -2t = $\frac{3}{8}$ t² – 2t× $\sqrt[]{}$ $\frac{3}{8}$ ×$\sqrt[]{}$ $\frac{8}{3}$ + $\frac{8}{3}$ -$\frac{8}{3}$
=( $\sqrt[]{}$ $\frac{3}{8}$ t-$\sqrt[]{}$ $\frac{8}{3}$ )² – $\frac{8}{3}$
ta có ($\sqrt[]{}$ $\frac{3}{8}$t -$\sqrt[]{}$ $\frac{8}{3}$ )² ≥0
⇒ ($\sqrt[]{}$ $\frac{3}{8}$ t-$\sqrt[]{}$ $\frac{8}{3}$ )² -$\frac{8}{3}$ $\geq$ $\frac{-8}{3}$
⇒ – ( $\sqrt[]{}$ $\frac{3}{8}$t -$\sqrt[]{}$ $\frac{8}{3}$ )² -$\frac{8}{3}$ $\leq$ ) ≤ $\frac{8}{3}$
⇒max = $\frac{8}{3}$