Tìm GTLN/GTNN ( nếu có):
A= 3x2-5x+7
B= (x-1)(x-3)=11
C= (x-3)2+(x-2)2
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a: =-x^2+6x-4
=-(x^2-6x+4)
=-(x^2-6x+9-5)
=-(x-3)^2+5<=5
Dấu = xảy ra khi x=3
b: =3(x^2-5/3x+7/3)
=3(x^2-2*x*5/6+25/36+59/36)
=3(x-5/6)^2+59/12>=59/12
Dấu = xảy ra khi x=5/6
c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)
\(=-\left[\left(\left|x-3\right|\right)^2-2\left|x-3\right|+1-1\right]\)
\(=-\left(\left|x-3\right|-1\right)^2+1< =1\)
Dấu = xảy ra khi x=4 hoặc x=2
a: \(f\left(x\right)=2x^2-7x+9\)
=>\(f'\left(x\right)=2\cdot2x-7=4x-7\)
Đặt f'(x)=0
=>\(4x-7=0\)
=>\(x=\dfrac{7}{4}\)
\(f\left(\dfrac{7}{4}\right)=2\cdot\left(\dfrac{7}{4}\right)^2-7\cdot\dfrac{7}{4}+9=\dfrac{23}{8}\)
\(f\left(-1\right)=2\left(-1\right)^2-7\cdot\left(-1\right)+9=18\)
\(f\left(4\right)=2\cdot4^2-7\cdot4+9=13\)
Vì \(f\left(\dfrac{7}{4}\right)< f\left(4\right)< f\left(-1\right)\)
nên \(f\left(x\right)_{max\left[-1;4\right]}=18;f\left(x\right)_{min\left[-1;4\right]}=\dfrac{23}{8}\)
b: \(f\left(x\right)=x^2+5x+3\)
=>\(f'\left(x\right)=2x+5\)
f'(x)=0
=>2x+5=0
=>2x=-5
=>\(x=-\dfrac{5}{2}\)
\(f\left(-\dfrac{5}{2}\right)=\left(-\dfrac{5}{2}\right)^2+5\cdot\dfrac{-5}{2}+3=\dfrac{25}{4}-\dfrac{25}{2}+3=-\dfrac{13}{4}\)
\(f\left(2\right)=2^2+5\cdot2+3=4+10+3=17\)
\(f\left(6\right)=6^2+5\cdot6+3=69\)
Vậy: \(f\left(x\right)_{max\left[2;6\right]}=69;f\left(x\right)_{min\left[2;6\right]}=-\dfrac{13}{4}\)
A = -4 - x2 + 6x = -(x2 - 6x + 9) + 5 = -(x - 3)2 + 5 \(\le\)5 \(\forall\) x
Dấu "=" xảy ra <=> x - 3 = 0 <=> x = 3
Vậy MaxA = 5 khi x = 3
F = (x - 1)(x - 3) + 11 = x2 - 4x + 3 + 11 = (x2 - 4x + 4) + 10 = (x - 2)2 + 10 \(\ge\)10 \(\forall\)x
Dấu "=" xảy ra <=> x - 2 = 0 <=> x = 2
Vậy MinF = 10 khi x = 2
B = 3x2 - 5x + 7 = 3(x2 - 5/3x + 25/36) + 59/12 = 3(x - 5/3)2 + 59/12 \(\ge\)59/12 \(\forall\)x
Dấu "=" xảy ra <=> x - 5/3 = 0 <=> x = 5/3
Vậy MinB = 59/12 khi x = 5/3
G = (x - 3)2 + (x - 2)2 = x2 - 6x + 9 + x2 - 4x + 4 = 2x2 - 10x + 13 = 2(x2 - 5x + 25/4) + 1/2 = 2(x - 5/2)2 + 1/2 \(\ge\)1/2 \(\forall\)x
Dấu "=" xảy ra <=> x - 5/2 = 0 <=> x = 5/2
Vậy MinG = 1/2 khi x = 5/2
A = 3x2 - 5x + 7
= 3( x2 - 5/3x + 25/36 ) + 59/12
= 3( x - 5/6 )2 + 59/12 ≥ 59/12 ∀ x
Dấu "=" xảy ra khi x = 5/6
=> MinA = 59/12 <=> x = 5/6
B mình chia thành hai trường hợp nhé ;-; trúng cái nào thì trúng :)
B = ( x - 1 )( x - 3 ) + 11
= x2 - 4x + 3 + 11
= ( x2 - 4x + 4 ) + 10
= ( x - 2 )2 + 10 ≥ 10 ∀ x
Dấu "=" xảy ra khi x = 2
=> MinB = 10 <=> x = 2
B = ( x - 1 )( x - 3 ) - 11
= x2 - 4x + 3 - 11
= ( x2 - 4x + 4 ) - 12
= ( x - 2 )2 - 12 ≥ -12 ∀ x
Dấu "=" xảy ra khi x = 2
=> MinB = -12 <=> x = 2
C = ( x - 3 )2 + ( x - 2 )2
= x2 - 6x + 9 + x2 - 4x + 4
= 2x2 - 10x + 13
= 2( x2 - 5x + 25/4 ) + 1/2
= 2( x - 5/2 )2 + 1/2 ≥ 1/2 ∀ x
Dấu "=" xảy ra khi x = 5/2
=> MinC = 1/2 <=> x = 5/2