Question

Find the absolute extrema of the function y=-x^2+3x-5, on [-2,1]

Answer 1

differentiate it twice, first

Answer 2

find the vertex, check the second coordinate of the vertex and the endpoints

Answer 3

don't differentiate at all

Answer 4

why twice?

Answer 5

@satellite73 it's easier

Answer 6

well, my teacher wants us to do it with differentiating.

Answer 7

vertex does not need calculus
first coordinate of the vertex is \(-\frac{b}{2a}\)

Answer 8

fine, take the derivative, set it equal to zero you will still get \(-\frac{b}{2a}\)

Answer 9

but calculus makes an oral calculation here , no, need to plot vertex...and see finally you used calculus

Answer 10

second deriative tells you nothing, since it is simply \(2a\) a constant, only tells you it is a parabola that faces down, which you know by looking

Answer 11

?

Answer 12

i used calc because it was requested. vertex is \((-\frac{b}{2a},f(-\frac{b}{2a}))\) calculus or no calculus

Answer 13

in any case you have to check 2x+y=-11,3x-4y=11
\[f(-2),f(1), f(\frac{3}{2})\] largest is max, smallest is min

Answer 14

y'=-2x+3

Answer 15

okay then ...you can guide better, i could have explained what that constant -2 means but sir, you carry on :)

Answer 16

i set it equal to 0 and got x=3/2

Answer 17

yes of course as \(a=1,b=3,-\frac{b}{2a}=\frac{3}{2}\) so now you are left to check
\[f(-2),f(1), f(\frac{3}{2})\]

Answer 18

yeah, i got that. do i put it into y' or y?

Answer 19

i.e. check the vertex and the endpoints of the interval to see which is largest and which is smallest

Answer 20

so, y?

Answer 21

you want the max and min of the function, put it in the function
if you replace \(x\) by \(\frac{3}{12}\) i the derivative you will get zero, because that is how you found it to begin with

Answer 22

yes use \(y\) not \(y'\) to evaluate

Answer 23

so i got 3/2, -15, and -3. my answer sheet says -15 and -3 are the answer but i dont understand why.

Answer 24

what is \(f(-2)\)?

Answer 25

-15

Answer 26

oh wait, i forgot to put 3/2 into the eqwuation

Answer 27

ok and \(f(1)=-3\) right?

Answer 28

ignore the \(\frac{3}{2}\) it is not in your interval

Answer 29

ohh! ok yeah i forgot about that. So how do I know which is the min and the max

Answer 30

interval is \([-2,1]\) and your vertex is at \(x=1.5\) so forget it, function is going up on your interval

Answer 31

not to be a wise guy, but the min is the min and the max is the max! you have \(-15\) at the left hand endpoint and \(-3\) at the right, evidently the min is \(-15\) and the max is \(-3\)

Answer 32

that makes sense. thank you!

Answer 33

don't forget minimum means minimum output, not input
yw