f'(x)=sqrt(x)*sin(x)
The first derivative of the f is given above. If f(0)=0, at what value of x does the function f attain its minimum value on the closed interval [0,10]?
Min value?
f'(x)=0=sqrt(x)*sin(x)
well at x=0 is solution.
sin(x)=0
x= n*PI is a solution up to
n<10/PI for n=1,2,3 test now second derivative...
f"(x)=cos(x)sqrtx + 1/2sqrtx *sinx
at x=0, is zero
at x= PI, is negative
at x=2PI, is positive
at x=3 pi, is negative
Minimum then by definition is at x=2PI, not counting the initial value at x=0
check my thinking.
looks good. f(x) has a min when f' changes from - to + (f bottoms out and starts back up)
That happens at x=2pi
Oh, finding the minimum value, huh? It's like searching for a needle in a haystack, except the needle is a min point and the haystack is the closed interval [0,10]. Good luck!
To find the minimum value, we'll need to find critical points first. We know that f'(x) is the derivative of f(x), so that's a good starting point. Let's find where f'(x) = 0 or where it doesn't exist.
Let's start by setting f'(x) = 0 and solving for x.
sqrt(x) * sin(x) = 0
Hmm, if you remember basic math, a product is zero if and only if at least one of the factors is zero. So that means either sqrt(x) = 0 or sin(x) = 0.
Well, sqrt(x) can never be zero on the interval [0,10], so we can cross that off the list.
Now, for sin(x) = 0, we know that sin(x) equals 0 at x = 0, π, 2π, …
Now we check whether any of those values for x are within the interval [0,10]. And voila! We find that x = 0 is indeed within the interval!
So, x = 0 is a critical point. Now we need to check the endpoints of the interval, [0,10], which are 0 and 10.
Since f(0) = 0, we have a point right there!
Now let's check f(10) to compare.
Unfortunately, finding f(10) is quite a task and way beyond my computational capabilities as a bot! You'll have to evaluate it yourself.
Anyway, now we've got x = 0 as a critical point and we have checked the endpoints as well. So, f(0) might be a potential minimum on the interval [0,10].
But remember, my friend, this is just a potential minima, not necessarily the actual minimum value. If you want to find the precise location of the minimum, keep on exploring!
To find the minimum value of the function f on the closed interval [0,10], we need to find the critical points and endpoints of the interval, and then check the value of the function at those points.
Step 1: Find the critical points.
To find the critical points, we need to solve f'(x) = 0.
f'(x) = √x * sin(x)
Setting √x * sin(x) = 0:
√x = 0 or sin(x) = 0
√x = 0 implies x = 0 as the only critical point.
Step 2: Evaluate the function at the critical point and endpoints of the interval.
We need to find the value of the function f(x) at x = 0, x = 10 (endpoints of the interval).
f(x) = √x * sin(x)
At x = 0: f(0) = √0 * sin(0) = 0
Now we can check the endpoints:
At x = 10: f(10) = √10 * sin(10)
Step 3: Compare the values to find the minimum.
Compare the values obtained at the critical point and the endpoints to find the minimum value.
f(0) = 0
f(10) = √10 * sin(10)
Since f(0) = 0 and f(10) is positive, the minimum value of f on the closed interval [0,10] occurs at x = 0.
Therefore, the function f attains its minimum value on the closed interval [0,10] at x = 0.
To find the minimum value of the function f on the closed interval [0,10] with the given derivative f'(x) = √(x) * sin(x), we can follow these steps:
1. Find the critical points of f(x) by setting the derivative f'(x) equal to zero and solving for x. In this case, we have:
√(x) * sin(x) = 0
Since sin(x) cannot be zero for any finite x, the critical point is when √(x) = 0. Solving this equation gives x = 0.
2. Evaluate the function f(x) at the critical points and the endpoints of the closed interval [0,10]. We have:
f(0) = 0 (Given condition)
f(10) = √(10) * sin(10)
3. Determine which of these values is the minimum value of the function f(x) on the closed interval [0,10].
Considering the interval [0,10], we need to compare the values of f(0), f(10), and any other critical points (if any) to determine the minimum value.
The minimum value could be at one of the endpoints or at a critical point. We evaluate the function at these points:
f(0) = 0
f(10) = √(10) * sin(10)
Comparing these values, we can see that f(0)=0 is the smallest value in the given interval. Therefore, the minimum value of the function f(x) on the closed interval [0,10] is 0, which occurs at x = 0.