Answer the following questions for the function
f(x) = sin^2(x/3)
defined on the interval [ -9.424778, 2.356194].
Rememer that you can enter pi for \pi as part of your answer.
f(x) is concave down on the interval ?
Please help, im REALLY stuck.
our domain is [-3π,3π/4]
f'(x) = 2(sin(x/3))(cos(x/3)
= sin (2x/3)
(I used sin 2A = 2sinAcosA)
f''(x) = (2/3)cos(2x/3)
finding the points of inflection ....
cos(2x/3) = 0
2x/3 = π/2 or 2x/3 = 3π/2
x = 3π/4 or x = 9π/4
but the period of cos(2x/3) = 2π/(2/3) = 3π
so within our domain we would have points of inflection at
x = 3π/4 or appr. 2.356 (the right side end point of domain)
x= 3π/4 - 3π = -9π/4 or appr. -7.1 (inside our domain) or
x = 9π/4 - 3π = -3π/4 or appr. -2.34 (inside our domain)
I use http://rechneronline.de/function-graphs/ to see my graphs.
Enter (sin(x/3))^2 into the function box, and change the "range x-axis" from -10 to 5.
You will see that the above results are correct.
so f(x) is concave down for -9π/4 < x < -3π/4
(remember concave down means f''(x) < 0 , check by picking any x in that spread and calculationg f''(x)
e.g. f''( -5) = (2/3)cos(-10.3) = -.65.. so concave down )
Thanks for posting that link. It's one of my favorite graphing sites.
To determine whether a function is concave down or concave up, we need to find the second derivative of the function and check its sign.
Given the function f(x) = sin^2(x/3), let's find its second derivative.
Step 1: Find the first derivative:
f'(x) = d/dx(sin^2(x/3))
To find the derivative of sin^2(u) with respect to u, we can use the chain rule:
d/du(sin^2(u)) = 2sin(u) * (d/dx(sin(u)))
Applying the chain rule to our function, we have:
f'(x) = 2sin(x/3) * (d/dx(sin(x/3)))
Step 2: Find the second derivative:
f''(x) = d/dx(f'(x))
= d/dx(2sin(x/3) * (d/dx(sin(x/3))))
= 2 * (d/dx(sin(x/3)) * (d/dx(sin(x/3))))
Using the chain rule again, we can find the derivative of sin(x/3):
d/dx(sin(x/3)) = (1/3) * cos(x/3)
Substituting this back into the second derivative, we get:
f''(x) = 2 * (d/dx(sin(x/3)) * (d/dx(sin(x/3))))
= 2 * (1/3) * cos(x/3) * (1/3) * cos(x/3)
= (2/9) * cos^2(x/3)
Now we need to determine the sign of f''(x) on the given interval [-9.424778, 2.356194].
On this interval, the cosine function is positive, so cos^2(x/3) is also positive.
Therefore, the second derivative f''(x) = (2/9) * cos^2(x/3) is always positive on the interval, indicating that the function f(x) = sin^2(x/3) is concave up on the entire interval [-9.424778, 2.356194].
Hence, f(x) is not concave down on the interval.
To determine whether the function f(x) = sin^2(x/3) is concave down on the interval [-9.424778, 2.356194], we can examine the second derivative of the function.
1. First, find the first derivative of f(x):
f'(x) = d/dx(sin^2(x/3))
Using the chain rule, we have:
f'(x) = 2*sin(x/3)*cos(x/3).
2. Next, find the second derivative of f(x):
f''(x) = d/dx(f'(x)).
Taking the derivative of f'(x), we get:
f''(x) = d/dx(2*sin(x/3)*cos(x/3))
Applying the chain rule again, we have:
f''(x) = 2*[cos(x/3)*cos(x/3) - sin(x/3)*sin(x/3)]
Simplifying the equation:
f''(x) = 2*(cos^2(x/3) - sin^2(x/3))
3. Determine the concavity of f(x):
Now, we need to analyze the sign of the second derivative within the given interval.
Since cos^2(x) is always positive and sin^2(x) is always positive, irrespective of the value of x, the sign of the second derivative only depends on the difference between cos^2(x/3) and sin^2(x/3).
Using the identity sin^2(x) + cos^2(x) = 1, we can substitute:
cos^2(x/3) = 1 - sin^2(x/3).
By substituting this back into the second derivative equation, we get:
f''(x) = 2*(1 - sin^2(x/3) - sin^2(x/3))
Simplifying further:
f''(x) = 2*(1 - 2*sin^2(x/3))
From the equation, it is apparent that 1 - 2*sin^2(x/3) is always less than or equal to 1. Hence, f''(x) is always positive or zero.
Therefore, the function f(x) = sin^2(x/3) is concave down (or flat) on the interval [-9.424778, 2.356194].