Question

Let f(x)=x^2+px+q. Find the values of p and q such that f(2)=2 is an extreme value of f on [0,4]. Is this value a max or min?

Answer 1

All I have so far is the derivative...

f ' (x)= 2x+p

Answer 2

Hint:

f(x)=x^2+px+q

is a parabola that opens upward (because the leading coefficient, which is 1, is positive)

Answer 3

Ok so this function will have an absolute min

Answer 4

good

Answer 5

the min occurs when f '(x) = 0

Answer 6

When finding the critical value, will the x-value include p.

This is my critical value: x=(-p/2)

Answer 7

or you can say because (2,2) is an extrema, you know that

f ' (2) = 0

Answer 8

oh ok, so p=-4

Answer 9

good

Answer 10

so

f(x)=x^2+px+q

turns into

f(x)=x^2-4x+q

Answer 11

how do you know (2,2) is an extrema?

Answer 12

it says

" ... f(2)=2 is an extreme value of f... "

Answer 13

oh dang, i wrote typed the problem wrong...
f(2)=3 is an extreme value of f

Answer 14

doesn't matter, things won't change too much

Answer 15

p is still -4

Answer 16

ok right because the x value is what is most important for finding p

Answer 17

yes

Answer 18

you've yet to use the y coordinate of the extrema

Answer 19

q=7

Answer 20

you got it

Answer 21

thanks for the help

Answer 22

So your function is

f(x) = x^2 - 4x + 7

You can verify by graphing

Answer 23

np