Calculus Problem (involes critical numbers)

In summary, the function h(x) = sin^2(x) + cos(x) has critical numbers at π/3, π, and 5π/3 within the interval 0 < x < 2π. These critical numbers can be found by setting the derivative of the function equal to zero and solving for x, which leads to the values of x = π/3, π, and 5π/3.
  • #1
Cod
325
4
The directions state to find any critical numbers of the function within the interval 0 < x < 2pi. The function is:

[tex]h(x) = sin^2(x) + cos(x)[/tex]

I've already found that the derivative is:

[tex]2sin(x)cos(x) - sin(x)[/tex]

I've set that equal to zero, and that's where I'm stuck. I guess this is more of an algebra question than anything. The answer I keep getting is:

[tex]tan(x)/sin(x) = 2[/tex]

However, I don't think its correct. So any help would be greatly appreciated.
 
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  • #2
[tex]2\sin{x}\cos{x} - \sin{x} = 0[/tex]
[tex]2\sin{x}\cos{x}= \sin{x}[/tex]
[tex]2\cos{x}=0[/tex]
[tex]\cos{x}=0[/tex]

...?

cookiemonster
 
  • #3
Ohhhhh, I'm an idiot. I was adding [tex]sin x[/tex] to both sides then dividing by [tex]cos x[/tex] to get [tex]tan x[/tex]. Well, I guess I was just looking over the obvious the whole time.

Thanks for the assistance.
 
  • #4
cod: Don't call yourself an idiot. Let's reserve that for cookiemonster! (Hey, it's not often I can do that!)

NO, 2sin x cos x= cos x does NOT immediately lead to
"2cos x= 0"- it leads to 2cos x= 1 !

In fact, the best way to do this is to factor the original form:
2sin x cos x- sin x= sin x(2 cos x- 1)= 0 so

either sin x= 0 or 2 cos x-1= 0. That is, either sin x= 0
or cos x= 1/2. You can get x itself from those.
 
  • #5
the answer for: Find any critical numbers of the function. (Enter your answers as a comma-separated list.)
(sin(x))2 + cos(x) 0 < x < 2π

is: Pi/3 , Pi, 5Pi/3 so you know
 

1. What are critical numbers in a calculus problem?

Critical numbers are values of a variable in a calculus problem where the derivative of the function is equal to zero or does not exist. They are important because they can help us identify maximum and minimum points on a graph.

2. How do you find critical numbers in a calculus problem?

To find critical numbers, you need to take the derivative of the function, set it equal to zero, and solve for the variable. The resulting values will be the critical numbers.

3. Why are critical numbers important in calculus?

Critical numbers help us find important information about a function, such as maximum and minimum points, inflection points, and points of intersection. They also help us determine the behavior of the function and its graph.

4. What is the difference between a local maximum and a global maximum?

A local maximum is the highest point on a specific interval of a function, while a global maximum is the highest point on the entire domain of the function. In other words, a global maximum is the largest possible value of the function, while a local maximum may not be the overall highest point.

5. How do you use critical numbers to solve optimization problems?

In optimization problems, we use critical numbers to find the minimum or maximum value of a function. We first find the critical numbers, and then test each value to see if it corresponds to a minimum or maximum point. The value that gives the highest or lowest output is the optimal solution to the problem.

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