Finding Concavity and Inflection Points of f(t) = 3cos^2(5t) on [-pi/5, pi/5]

[smith5029]

hey guys I'm having trouble on these things.

ok here it is f(t):3cos^2(5t), the domain is [-pi/5,pi/5].

I have to find the concavity and where the curve increases and decreases. I took the second derivative and got the inflection point at .157.

I have no idea where i am going wrong but i am. please help me out.

 

[HallsofIvy]

Well, let's take a look at it. f(t)= 3 cos²(5t) so
f'(t)= 6 cos(5t)(-sin(5t))(5)= -30 cos(5t)sin(5t).
f"(t)= (-30)(- 5 sin(5t)sin(5t)- 30 cos(5t)(5cos(5t)= -150(cos²(5t)- sin²(5t))= -150cos(10t). That will be 0 when 10t is an odd multiple of pi/2: that is, when pi= pi/10, -pi/10, 3pi/10, -3pi/10. None of those is equal to 0.157 (pi/20?). Since the second derivative can change sign only at those points, for t between -pi/10 and pi/10, f"(t) has the same sign as f"(0)= -150. f(t) is concave downward for -pi/10< t< pi/10, positive for pi/10< t< 3pi/10, etc.

Of course, the curve increases and decreases where f'(t) is positive and negative, respectively.

 

[M98Ranger]

Hi there,

No worries, finding concavity and inflection points can be tricky at times. Let's break down the steps for finding these points for the given function f(t) = 3cos^2(5t).

1. Find the first derivative of the function f(t) using the power rule and chain rule. The first derivative will give you information about the increasing and decreasing intervals of the function.

f'(t) = -30sin(5t)cos(5t)

2. Set the first derivative equal to 0 and solve for t to find the critical points:

-30sin(5t)cos(5t) = 0
sin(5t) = 0 or cos(5t) = 0

This gives us critical points at t = 0, t = pi/10, and t = 3pi/10.

3. Now, we need to determine the concavity of the function. To do this, we will take the second derivative of f(t).

f''(t) = -150cos(10t)

4. Set the second derivative equal to 0 and solve for t to find the inflection points:

-150cos(10t) = 0
cos(10t) = 0

This gives us inflection points at t = pi/20 and t = 3pi/20.

5. Finally, we can create a number line to determine the intervals of concavity and the inflection points. Use the critical points and inflection points as your guide and test points within each interval to determine the concavity.

Interval: [-pi/5, 0]
Test point: -pi/10
f''(-pi/10) = -150cos(10(-pi/10)) = -150cos(-pi) = 150
Since the second derivative is positive, the function is concave up in this interval.

Interval: [0, pi/10]
Test point: pi/20
f''(pi/20) = -150cos(10(pi/20)) = -150cos(pi/2) = 0
Since the second derivative is 0, we cannot determine the concavity in this interval. However, we know that there is an inflection point at t = pi/20.

Interval: [pi/10, 3pi/10]
Test point: 3pi/20
f''