What are the critical points of the given function? f(x)=(1/3)x^2 +2cosx

Question

f(x)=(1/3)x^2 +2cosx

I'm stuck on this question because I feel like there is no answer. To use newtons law you need to use roots or when x=0 but there are no points on the x axis. Can someone please explain this to me and guide me towards the answer.

Thanks.

Answer 1

F'(x) = 2/3 (x) - 2 sinx

To find out when the derivative is zero graph y = 2/3 (x) - 2sinx and determine where the graph crosses the x-axis, i.e. the roots of the derivative. These will be your critical numbers. I did the same I found 3 critical numbers (-2.279, 0, and 2.279). You can apply the first derivative test to determine if those critical numbers create local max/min for the original function.

Answer 2

To find critical points, take the derivative and set it equal to zero.

if f(x) = (1/3)x² + 2cos(x), then

f '(x) = 2/3x - 2sin(x) then setting this equal to zero, we get:

2/3x - 2sin(x) = 0 or

1/3x - sin(x) = 0

I thought for a bit and didn't see a way to solve this algebraically so I resorted to using my calculator. Plugging in y = 1/3x - sin(x) and looking for x-intercepts, I found three of them:

x = -2.27886266

x = 0

x = 2.27886266

These are the critical points.