Hello. I have a few questions from my study guide, that I need to know to study for my test.
1) x^(2/3)*((5/2)-(x)
a) determine the ordered pairs of the local extrema of the function. Use the second derivative test.
B) determine the ordered pairs of all inflection points of the function.
2) Verify that the hypotheses for the Mean Value Theorem hold and find all the values of C that satisfy it's conclusion
F(x)= Cosx-Sin x [0,2pi)
y = x^(2/3) * ((5/2)-(x)
y' = (2/3)x^(-1/3)*(5/2 - x) + x^(2/3)(-1)
= 5(1-x) / (3*x^(1/3))
So, y'=0 when x=1
y" = -5(2x+1) / (9x^(4/3))
y"(1) < 0, so f(1) is a maximum
See the graph at
http://www.wolframalpha.com/input/?i=+%28x^2%29^%281%2F3%29+*+%28%285%2F2%29-%28x%29
Hello! I can help you with your questions.
1) To find the ordered pairs of the local extrema of the function, you need to follow these steps:
- First, find the critical points of the function by finding where the derivative is equal to zero or undefined. In this case, take the derivative of the given function.
So, f'(x) = (2/3)x^(-1/3) * (5/2) - 1 = (10/6)x^(-1/3) - 1 = (10-6x^(1/3))/6x^(1/3)
- Now, set f'(x) equal to zero and solve for x to find the critical points. In this case, set (10-6x^(1/3))/6x^(1/3) = 0
Simplifying, we get 10 - 6x^(1/3) = 0, or 6x^(1/3) = 10.
Solving for x, we get x = (10/6)^3 = 125/36.
So, the critical point is x = 125/36.
- Next, to determine the nature of the extrema, we need to evaluate the second derivative of the function at the critical point. Find f''(x) by taking the derivative of f'(x).
f''(x) = (10/6) * (1/3) * x^(-4/3) = 5/(18x^(4/3)).
- Finally, substitute the critical point, x = 125/36, into f''(x) and determine the sign of f''(x). If f''(x) is positive, it corresponds to a local minimum, and if f''(x) is negative, it corresponds to a local maximum.
Substituting x = 125/36 into f''(x), we get f''(125/36) = 5/(18 * (125/36)^(4/3))
Simplifying, we get f''(125/36) = 5/(18 * (5/4)^4)
Compute the final value to determine the sign.
Once you determine the sign of f''(x) at the critical point, you can find the ordered pairs of the local extrema.
For part (b) to find the ordered pairs of all inflection points, follow these steps:
- Find the points where the second derivative is equal to zero. Set f''(x) = 0 and solve for x.
- Once you find the points x where f''(x) = 0, substitute these values into the original function to find the corresponding y-coordinate or function values.
2) To verify if the hypotheses for the Mean Value Theorem hold for the function f(x) = cos(x) - sin(x) on the interval [0, 2π), follow these steps:
- The hypotheses for the Mean Value Theorem require that the function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b).
In this case, the function f(x) = cos(x) - sin(x) is the difference of two trigonometric functions, and since cosine and sine functions are continuous and differentiable all over the real line, the function f(x) is also continuous and differentiable on the interval [0, 2π).
- Next, to find all values of C that satisfy the conclusion of the Mean Value Theorem, which states that there exists a c in the open interval (a, b) where f'(c) = (f(b) - f(a))/(b - a), perform the following steps:
Find the derivative of f(x), which is f'(x).
f'(x) = -sin(x) - cos(x)
Substitute the values a = 0 and b = 2π into f(x) and f'(x). Compute the values of f(2π) - f(0) and 2π - 0.
f(2π) = cos(2π) - sin(2π)
f(0) = cos(0) - sin(0)
2π - 0 = 2π
Finally, solve the equation f'(c) = (f(2π) - f(0))/(2π - 0) for c.
-f'(c) = (f(2π) - f(0))/(2π - 0)
-sin(c) - cos(c) = (cos(2π) - sin(2π))/2π
By simplifying the equation and solving for c, you will find the values of c that satisfy the conclusion of the Mean Value Theorem.
I hope this explanation helps you understand how to solve these problems. Let me know if you have any further questions!