Question

tìm GTLN của C= \(3x-x^2\)

Answer 1

ta có :C=3x-x\(^2\)

=-(x\(^2\)-3x)

C=3x-x\(^{^{ }2}\)

=-(x\(^2\)-3x)

=-[(x\(^{^{ }2}\)-2x\(\dfrac{3}{2}\)+\(\dfrac{9}{4}\))-\(\dfrac{9}{4}\)]

=-(x-\(\dfrac{3}{2}\))\(^2\)+\(\dfrac{9}{4}\)

vì -(x-\(\dfrac{3}{2}\))\(^{^{ }2}\)\(\le\)0 với mọi x

do đó max C=\(\dfrac{9}{4}\)

Answer 2

C= $3x-x^2$

= -x²+3x+\(\dfrac{9}{4}-\dfrac{9}{4}\)

= \(\left(-x^2+3x-\dfrac{9}{4}\right)+\dfrac{9}{4}\)

= - \(\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{9}{4}\)

= -\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{9}{4}\)

do \(-\left(x-\dfrac{3}{2}\right)^2\le0\forall X\)

=> -\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

=> C ≤ \(\dfrac{9}{4}\)

vậy GTLN C =\(\dfrac{9}{4}\)khi x-\(\dfrac{3}{2}=0\) => x=\(\dfrac{3}{2}\)