Question

Concavity

Answer 1

F(x) = cos(x)-x ; on the interval (0,2pi)

Find Inflection point:
Concave up:
Concave down:

Answer 2

Can you calculate F'(x) and then F''(x) ? That's the first step.

Answer 3

Yea
\[f'(x) = -sin(x)-1\]\[f''(x) = -cos(x)\]

Answer 4

Derivatives look good. What's step #2 ?

Answer 5

Well if it's like these other ones I've done, set it equal to zero

Answer 6

Yes. Set f''(x) = 0 and solve for x on your (0,2pi) interval. Any answers on that interval will be inflection points

Answer 7

So , x = -cos ?

Answer 8

No, f''(x) = 0 = -cos x. Which angles on (0,2pi) will result in -cos x to be 0. There are two values.

Answer 9

pi/2 and 3pi/2

Answer 10

Yes, those will be the points of inflection on f(x). Now, we have to do the Second Derivative Test to determine concavity of f(x). Do you know how to do that?

Answer 11

Yes

Answer 12

Sweet, there ya go.

Answer 13

On this interval, I got downward from 0 to 2pi. Can you check this please

Answer 14

Not quite. The concavity will change 3 times since there are 2 inflection points.