loc max, loc min; glob max, glob min for f(x) = sin^2x - cos

[calculus 1983]

Given: f(x) = sin^2x - cos
on the closed interval [0, pi]

a) find: loc max, loc min
b) find: glob max, glob min

a) loc max and loc min on a closed interval:
- 1st deriv test with endpoints

f(x) = sin^2x - cos x
rewrite = (sinx)^2 - cosx

f'(x) = 2(u) x du/dx (or the derivative of you with respect to x) - (-sinx)
f'(x) = 2(sinx) x (cosx) + sinx
f'(x) = 2sinxcosx +sinx
factor: f'(x) = sinx (2cosx + 1) set it equal to 0

sin x = 0

x = 0
x = pi

(Do you get x=0 and x=pi by plugging in y = sinx in the calculator and finding out where sinx is 0?)

Set 2cosx + 1 = 0
2cosx = -1
cosx = -1/2

put it in the calculator as cos^-1 (1/2) which gives you approx. 1.05 which is your reference angle

(Also, why do you plug -1/2 as 1/2 in the calculator?)

This is what i am most confused about i know the All Students Take Calc ...and how All is 1st quadrant, Sin is 2nd, Tan is 3rd, and Cosine is 4th
but why do i plug it in the second quadrant, how do you determine that?
(and the second quadrant is pi - x .... so it is reference angle = pi - x ..or 1.05 = pi - x ; -2.09 = -x ; 2.09 = x)

 

[skeeter]

have you not studied trigonometry at all?

this problem was meant to be done by hand ... i.e., no calculator.

f(x) = sin²x - cosx

f'(x) = 2sinxcosx + sinx

2sinxcosx + sinx = 0
sinx(2cosx + 1) = 0

sinx = 0 ... you're expected to know that sinx = 0 at x = 0, x = pi, and x = 2pi

cosx = -1/2 ... you're expected to know that cosx = -1/2 at x = 2pi/3 and x = 4pi/3

recommend you learn the unit circle

you have critical values of x at x = 0, x = 2pi/3, x = pi, x = 4pi/3, and x = 2pi

for values of x between 0 and 2pi/3, f'(x) > 0 ... f(x) is increasing on that interval.

for values of x between 2pi/3 and pi, f'(x) < 0 ... f(x) is decreasing on that interval.

for values of x between pi and 4pi/3, f'(x) > 0 ... f(x) is increasing on that interval.

for values of x between 4pi/3 and 2pi, f'(x) < 0 ... f(x) is decreasing on that interval.

you have maximums at x = 2pi/3 and x = 4pi/3

you have mimimums at x = 0, x = pi, and x = 2pi

determine the value of f(x) at each x-value listed above to determine which are relative and which are absolute extrema.