Find the value(s) of x for the following. (Enter your answers as a comma-separated list.)
f(x) = 6 sin^2 x − 6 cos x,
and the interval is 0 is less than or equal to x and x is less than or equal to pi.
(a) f(x) has a local maximum or local minimum.
local maximum: ??
local minimum : ??
(b) f(x)has a global maximum or global minimum.
global max: ??
global min: ??
I do not know how to solve this problem. I understand that I'm supposed to find the derivative first but I don't even know how to take the derivative of sin^2x. Please help me solve this problem!
Sorry I think parts of the answer you gave are not visible on the page.
for the derivative, remember the chain rule. In this case, u=sinx and f(u) = u^2, so f'(x) = 2u u', as below.
f = 6 sin^2 x − 6 cos x
f' = 12sinx cosx + 6sinx
(a) max/min where f'=0
6sinx(2cosx+1) = 0
f" = 6cosx + 6cos(2x)
max if f" < 0
min if f" > 0
(b) check extrema for greatest values. Since f is periodic, there won't be many to check.
So the critical points are 0, pi, and 2pi/3 but I don't know where to go from there..?
To find the value(s) of x for the given function f(x) = 6 sin^2(x) - 6 cos(x), we need to determine whether f(x) has local maximum or minimum values and global maximum or minimum values within the interval [0, π].
(a) To find the local maximum or minimum values of f(x), we need to find the critical points. In order to do that, we first need to take the derivative of f(x).
To calculate the derivative of sin^2(x), you can apply the chain rule.
Let's start with finding the derivative of sin^2(x):
Let u = sin(x), then sin^2(x) = u^2.
Applying the chain rule, we have:
d(sin^2(x))/dx = d(u^2)/dx = 2u * du/dx
Now, we need to find du/dx. Taking the derivative of u = sin(x) with respect to x gives:
du/dx = cos(x)
Substituting this back into the previous expression, we have:
d(sin^2(x))/dx = 2u * (du/dx) = 2sin(x) * cos(x)
Now, let's find the derivative of -6cos(x):
d(-6cos(x))/dx = -6 * d(cos(x))/dx = -6 * (-sin(x)) = 6sin(x)
Finally, we can find the derivative of f(x):
f'(x) = 6sin(x) * 2cos(x) + 6sin(x) = 12sin(x) * cos(x) + 6sin(x)
To find the critical points, we set f'(x) equal to zero:
12sin(x) * cos(x) + 6sin(x) = 0
Factoring out sin(x), we get:
sin(x) * (12cos(x) + 6) = 0
Now, we have two possibilities for f'(x) to be zero:
1) sin(x) = 0
2) 12cos(x) + 6 = 0
To solve these equations, let's analyze each one:
1) sin(x) = 0
This equation is satisfied when x = 0 or x = π.
2) 12cos(x) + 6 = 0
Subtracting 6 from both sides, we have:
12cos(x) = -6
Dividing both sides by 12:
cos(x) = -1/2
This equation is satisfied when x = 2π/3 or x = 4π/3.
So, the critical points (values of x) where f'(x) = 0 are x = 0, x = π, x = 2π/3, and x = 4π/3.
Now, we need to determine whether these critical points are local maximum or minimum. To do that, we can analyze the second derivative of f(x).
First, let's find the second derivative of f(x):
f''(x) = d/dx (12sin(x) * cos(x) + 6sin(x))
Using the product rule, we have:
f''(x) = (12cos^2(x) - 12sin^2(x) + 6cos(x))
Now, let's evaluate the second derivative at each of the critical points:
For x = 0:
f''(0) = (12cos^2(0) - 12sin^2(0) + 6cos(0))
= 12(1) - 12(0) + 6(1)
= 12 + 6
= 18
For x = π:
f''(π) = (12cos^2(π) - 12sin^2(π) + 6cos(π))
= 12(-1) - 12(0) + 6(-1)
= -12 - 6
= -18
For x = 2π/3:
f''(2π/3) = (12cos^2(2π/3) - 12sin^2(2π/3) + 6cos(2π/3))
= 12(-1/2)^2 - 12(√3/2)^2 + 6(-1/2)
= 12(1/4) - 12(3/4) - 3
= 3 - 9 - 3
= -9
For x = 4π/3:
f''(4π/3) = (12cos^2(4π/3) - 12sin^2(4π/3) + 6cos(4π/3))
= 12(-1/2)^2 - 12(-√3/2)^2 + 6(-1/2)
= 12(1/4) - 12(3/4) - 3
= 3 - 9 - 3
= -9
Now, we can conclude:
A local maximum occurs when f''(x) < 0, and a local minimum occurs when f''(x) > 0.
(a) f(x) has a local maximum or local minimum:
- f(x) has a local maximum at x = π because f''(π) < 0.
- f(x) has a local minimum at x = 0, x = 2π/3, and x = 4π/3 because f''(0), f''(2π/3), f''(4π/3) > 0.
(b) To find the global maximum or minimum of f(x), we need to evaluate f(x) at the endpoints of the interval, which are x = 0 and x = π.
Calculate f(0):
f(0) = 6sin^2(0) - 6cos(0) = 0 - 6(1) = -6
Calculate f(π):
f(π) = 6sin^2(π) - 6cos(π) = 0 - 6(-1) = 6
Therefore, the global maximum of f(x) is 6 at x = π, and the global minimum of f(x) is -6 at x = 0.
In summary:
(a) Local maximum: x = π
Local minimum: x = 0, x = 2π/3, x = 4π/3
(b) Global maximum: 6 at x = π
Global minimum: -6 at x = 0
If then
That's all I can do for you since I don't know whether you mean or .
Either way, there will be two answers depending on the signs of and
John
My calculator said it, I believe it, that settles it