Question

determine the abs extreme values of the function f(X)=sinx - cosx+6 on the interval 0 to 2pi

Answer 1

i got the derivative but dont know how to solve for x.

Answer 2

You have to use extreme value theorem.

Answer 3

because you're looking for absolute extremes, you'll need to look at the local max/min and the values at the endpoints. then you must compare those values which is the most extreme.

Answer 4

what's your derivative?

Answer 5

f'(x)=cosx+sinx

Answer 6

ok, set f'(x) = 0.... and solve for x.
is this where you're stuck?

Answer 7

yes

Answer 8

i get -tanx=0

Answer 9

\[\large cosx+sinx=0\rightarrow sinx=-cosx\rightarrow \frac{sinx}{cosx}=-1\rightarrow tanx=-1 \]

Answer 10

where is tangent negative? what quadrants?

Answer 11

kk got it. 2nd and 4th

Answer 12

ok... what's x in those quadrants?

Answer 13

i got it now.

Answer 14

thanks.

Answer 15

ok... we done then?

Answer 16

yw... :)

Answer 17

1 more.

Answer 18

y=cos^3x

Answer 19

wht u need done here?

Answer 20

y'

Answer 21

and equation of tanget at 3pi.

Answer 22

chain rule...
\[\large y=cos^3x=(cosx)^3 \]
\[\large y'=3(cosx)^2(-sinx)=-3sinxcos^2x \]

Answer 23

3pi = pi so
\[\large y'|_{x=pi}=-3sin(\pi)cos^2(\pi)=-3(0)(-1)^2=0 \]
this is the slope of the tangent line at x=pi

Answer 24

ok.

Answer 25

now i know the slope.. which is 0, how do i find the b value? (y=mx+b)