Question

Morningg(: For the function f(x) = 2cos x, apply the Second Derivative Test to identify the values of x at which the function has local extrema in the interval [-3, 3]. Maximum and minimum please (:

Answer 1

Find second derivative

Answer 2

Find the first derivative.

Find the zeros of the first derivative.

Find the second derivative.

Evaluate the second derivative for values that were zeros of the first derivative.

If the result is positive, then it is a minimum.

If the result is negative, then it is a maximum.

Answer 3

Also test the boundaries

Answer 4

Yes, that too. :)

Answer 5

OKay the 1st derivative is -2sinx

Answer 6

Right so now in the interval solve for -2sinx = 0

Answer 7

In the interval [-3, 3], what are the zeros of sin x ?

Answer 8

OKay the 1st derivative is -2sinx

Answer 9

OKay the 1st derivative is -2sinx

Answer 10

I thought you said second derivative though. Im confused :/

Answer 11

The places were there are maxima or minima are zeros of the first derivative.

Answer 12

Then the second derivative will tell you which they are.

Answer 13

sin x = 0 when x = 0, pi, 2pi, 3pi, etc.

So inside [-3, 3], the only value is 0.

So now take the second derivative

f''(x) = -2cos x

f''(0) = -2(1) = -2 Since this is negative then the function has a local maximum at x = 0

Answer 14

Im so lost. What are the values of pi, 2pi,3pi for ?

Answer 15

Because it is a periodic function. It has zeros at all those values, but they are all not inside the [-3, 3] interval.

Answer 16

http://bit.ly/neJs5M

Answer 17

ohhh. Okay thanks(: