Is this the correct answers for these questions
Verify the means value theorem holds on the interval shown. Then, find the value c such that f'(c)=(f(b)-f(a))/(b-a)
b.f(x)=x^3=x-4 on [-2,3] c= square root 7/3
c. f(x)= x^3 on [-1,2] c= square root 1
d. f(x)= Sqr. root of x on[0,4]c= square root 1
b.f(x)=x^3=x-4
Typo, makes no sense
c. f(2)=8
f(-1)=-1
so
9/3 average slope =3
f' =3x^2
3 x^2 =3
x= sqrt(1) which is 1 !!!!
but also -1
d f(4) = 2
f(0) = 0
so slope = 2/4 = 1/2
f' = (1/2)x^-1/2 = 1/2
x = 1 again
I'll do (b) to show the way. Of course, you probably already know the way...
Assuming you meant
f(x) = x^3+x-4,
f(-2) = -14
f(3) = 26
slope of secant is 40/5 = 8
f' = 3x^2+1
3x^2+1 = 8
3x^2 = 7
...
clearly, √(7/3) is in [-2,3]
To verify if the mean value theorem holds for a given function on a specific interval, you need to check two conditions:
1. Continuity: The function should be continuous on the closed interval [a, b].
2. Differentiability: The function should be differentiable on the open interval (a, b).
Now let's verify if the mean value theorem holds for each given function and interval:
a. f(x) = x^3 - x - 4 on the interval [-2, 3]
To verify continuity, we need to check if the function is continuous on the closed interval [-2, 3]. Since polynomial functions are continuous everywhere, it is continuous on the interval [-2, 3].
To verify differentiability, we need to check if the function is differentiable on the open interval (-2, 3). Taking the derivative of f(x), we have f'(x) = 3x^2 - 1. The derivative is defined and continuous for all real numbers, so it is differentiable on the open interval (-2, 3).
Since the function satisfies both conditions, we can apply the mean value theorem. According to the mean value theorem, there exists a value c in the open interval (-2, 3) such that f'(c) = (f(3) - f(-2))/(3 - (-2)).
Calculating f'(x) = 3x^2 - 1:
f'(x) = 3x^2 - 1
Calculating f(3) and f(-2):
f(3) = (3)^3 - (3) - 4 = 27 - 3 - 4 = 20
f(-2) = (-2)^3 - (-2) - 4 = -8 + 2 - 4 = -10
Calculating (f(3) - f(-2))/(3 - (-2)):
(f(3) - f(-2))/(3 - (-2)) = (20 - (-10))/(3 + 2) = 30/5 = 6
So, to find the value c, we need to solve f'(c) = 6.
f'(c) = 3c^2 - 1 = 6
3c^2 = 7
c^2 = 7/3
c = sqrt(7/3)
Therefore, the value of c such that f'(c) = (f(3) - f(-2))/(3 - (-2)) is c = sqrt(7/3).
Hence, the correct answer is option b.
Please note that for the other options (c and d), the calculations were not provided. If you provide the necessary information, I can help you solve them as well.