Question

Find the smallest possible value of \( 2x^2 - 3x + 4 \)

Answer 1

find the derivative then set it equal to zero

Answer 2

Or complete the square and read off the minimum.

Answer 3

4x - 3 = 0

x = 3/4

then plug this in the equation

Answer 4

21/8

Answer 5

2(x^2 - 3/2 x + 9/4) + 4 - 9/4
= 2(x - 3/2)^2 + 4 - 9/4

is it 4 - 9/4 ?

Answer 6

I think I did a tiny mistake

Answer 7

you need to subtract 2(9/4) from the outside of the parenthesis to balance the equation.

Answer 8

okay... Ill try again

Answer 9

you have everything else right, it just needs to be:\[4-\frac{18}{4}=-\frac{1}{2}\]

Answer 10

2x^2 - 3x + 4
= 2 (x^2 - 3/2 x + 2)
= 2(x^2 - 3/2 x) + 4

why am I even struggling with this :-)

Answer 11

but the answer book says the smallest value is actually 23/8

Answer 12

it should be 2(x - 3/4)^2 + 4 - 9/8

Answer 13

oh I see my mistake now

Answer 14

'and at that moment, Agdgdgdgwngo was enlightened'

Answer 15

yeah i just saw it too >.< there should be a 9/16's inside.

Answer 16

Yes,
2x^2−3x+4 = 2(x^2 - 3/2.x +2)
= 2[ (x - 3/4)^2 + 23/16 ]