1. ## concavity

If n is a positive integer, how many times does the function f(x) = x^2 + 5cosx change concavity in the interval 0 <= x <= 2*pi*n?

A) 0
B) 1
C) 2
D) n
E) 2n

2. Originally Posted by DINOCALC09
If n is a positive integer, how many times does the function f(x) = x^2 + 5cosx change concavity in the interval 0 <= x <= 2*pi*n?

A) 0
B) 1
C) 2
D) n
E) 2n
The concavity of a function is its second derivative.

So find the second derivative and set it equal to 0. (Is this last step even possible?)

-Dan

3. 1st derivative = 2x - 5sinx
2nd derivative = 2 - 5cosx

set 2nd derivative = 0

so:

2 - 5cosx = 0
cosx = -2 / -5
x = arccos(-2/5)

This is only satisfied once?

So the answer should be (B) ?

4. or is the answer E?

thats what my friend got. we still lack a definent answer though.

5. Originally Posted by DINOCALC09
If n is a positive integer, how many times does the function f(x) = x^2 + 5cosx change concavity in the interval 0 <= x <= 2*pi*n?

A) 0
B) 1
C) 2
D) n
E) 2n
Look for points of inflexion (by solving f''(x) = 0) over the given interval.

6. i believe i did that up above... not sure if i did it right or not, but yeah =/

7. Originally Posted by DINOCALC09
1st derivative = 2x - 5sinx
2nd derivative = 2 - 5cosx

set 2nd derivative = 0

so:

2 - 5cosx = 0
cosx = -2 / -5
x = arccos(-2/5) Mr F says: This should be arccos(2/5).

This is only satisfied once? Mr F says: There are two solutions over the interval 0 < x < 2pi. To see this, draw the graphs of y = cos x and y = 2/5 over 0 < x < 2pi. You should see two intersection points ......

And since the period of cos x is 2pi, you'll have two solutions over every interval of size 2 pi ..... So four solutions over 0 < x < 4pi, six solutions over 0 < x < 6pi ......, 2n solutions over 0 < x< 2npi

So the answer should be (B) ?