Question

max & min

Answer 1

find the max of f(x)=8x^2 + 5x + 6
on the interval [-20,20].

Answer 2

it will be max? or min?

Answer 3

my hw is on max and mins but this problem wants just the max

Answer 4

it can't be max though.
Coz the value of "a" is +ve.
which means a min point.

Answer 5

o its supposed to be -8x^2+5x+6...
forgot the negative

Answer 6

Yes!
Now to find the dy/dx. then equal that to zero.
Solve for x.

Answer 7

i just learned this yesterday!

Answer 8

so -8x^2+5x+6=0
and solve to get the max?

Answer 9

min.

Answer 10

\[\large \frac{d}{dx} (-8x^2 +5x+6) = 5 - 16x\]

Answer 11

After we find the derivative, set the equation equal to zero.

5 - 16x = 0
-16x = -5
x = 5/16

Answer 12

i see

Answer 13

Now we have the "x" cordinate of the vertex.

Answer 14

where do i go from here to get the max?

Answer 15

Now, after u have the value of "x". Substitute it in the original equation to get the value if y.

Answer 16

of*

Answer 17

oh yeah duh

Answer 18

thanks saifooo

Answer 19

Welcome! :D

Answer 20

i cant seem to find the min of a similar problem...
g(t)=t^3-9t+5 on interval [-1.5, 3]

Answer 21

why isnt it 3?