Question

Find the absolute maximums and minimums of f on the given closed interval and state where these values occur.

f(x) = sinx-cosx

Answer 1

The closed interval is 0,pi

Answer 2

I can't do it manually but bu graphing it on a ti-84 between 0 and pi I get max y of 1.4142136 @ x of 2.3561932 and min y of -1 at x of 0.

Answer 3

Did you use a graphing calculator?

Answer 4

yes
also, you may recognize the 1.414 ar sqrt2

Answer 5

Alright well thanks for helping. You've got the best response so far :)

Answer 6

is that a calc 1 question?

Answer 7

AP Calc AB yeah

Answer 8

write as a single function of sine, then it will be much easier

Answer 9

sinx = cosx ? ?

Answer 10

take the derivative, set = 0 ,

Answer 11

well taking the derivative of the original equation,

0 = cosx + sinx

Answer 12

\[a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\] where
\[\cos(\theta)=\frac{a}{\sqrt{a^2+b^2}}, \sin(\theta)=\frac{b}{\sqrt{a^2+b^2}}\]

Answer 13

divide by cos x

Answer 14

in your case \(a=1,b=-1\) and so it is
\[\sqrt{2}\sin(x-\frac{\pi}{4})\]

Answer 15

0=sinx/cosx = tanx

Answer 16

then no calc needed
you see the answer from your eyeballs

Answer 17

0 = cosx + sinx
0= 1 + tan x

Answer 18

true sat, but he's taking an AP calc test

Answer 19

the largest
\[\sqrt2\sin(x-\frac{\pi}{4})\] can be is \(\sqrt{2}\) and the min is \(-\sqrt{2}\)

Answer 20

true, but isn't is more obvious by symmetry that the largest \(\sin(x)-\cos(x)\) it can be is if \(\sin(x)=-\cos(x)\) ?

Answer 21

sat, I am a little confused. What did you do after you found the derivative.

Answer 22

i never found the derivative
i used the identity
\[\sin(x)-\cos(x)=\sqrt2\sin(x-\frac{\pi}{4})\]

Answer 23

the amplitude is \(\sqrt{2}\) so that is the max

Answer 24

Ok i see the problem, and maybe it is because my teacher does it slightly diffferent. How did you get sqrt2sin(x-pi/4)

Answer 25

i gotta run, but i wrote all the steps for finding it above

Answer 26

I'll look above. Thanks for the help

Answer 27

yw

Answer 28

just in case you want to use calc:
d (sinx-cosx) = cos x + sin x = 0
1 + tan x = 0
tan x = -1
x = 135º and 315º (or 3pi/4 and -pi/4 )
in the interval 0 to pi, we only have 3pi/4
sin 3pi/4 = sqr(2)/2 cos 3pi/4 = - sqr(2)/2
sin - cos = sqr(2)

to find the min, test the boundaries 0 and pi for the min value

Answer 29

So that's where the sqrt2 came from. This looks more like what my teacher was showing me. I like this explanation better thank you.

Answer 30

sat did not use calculus... more a trig approach. It is a technique people often use when solving differential equations.