Verify that the function f(x)=3x^2+2x=5 satifies the hypotheses of the Mean Value Theorem on th interval[-1,1]

Question

Verify that the function f(x)= 3x^2 + 2x = 5 satifies the hypotheses of the Mean Value Theorem on th interval [-1,1]. Then find all number(s) c that satisfy the conclusion of the Mean Value Theorem.

Answer 1

The mean value theorem states that the slope of the function at some point in the interval will equal the average slope over the interval (or the slope of the line connecting the two endpoints of the interval).

So...

f(-1) = 3 - 2 = 1
f(1) = 3 + 2 = 5

The slope of the line connecting those two points is:

delta-y/delta-x =
(f(x2) - f(x1)) / (x2 - x1) =
(5-1) / (1 - (-1)) =
4/2 =
2

Now you have to find the places where the slope of f(x) is equal to 2

f'(x) =
d(3x^2+2x)/dx =
6x + 2

You need to solve for x:

f'(x) = (f(x2) - f(x1)) / (x2 - x1)
6x + 2 = 2
6x = 0
x = 0

The number(s) c include only x=0

Answer 2

Answer: c = 0
------------------
Mean Value Theroem: if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that
d[f(c)]dx = (f(b) - f(a))/(b - a)

f(x)= 3x^2 + 2x + 5 --> your problem stated "f(x)= 3x^2 + 2x = 5" so I took a chance that the "=" should have been a "+"

d[f(x)]/dx = 6x + 2, d[f(c)]/dx = 6c + 2

f(-1) = 3((-1)^2) + 2(-1) + 5 = 3 - 2 + 5 = 6
f(1) = 3((1)^2) + 2(1) + 5 = 3 + 2 + 5 = 10

6c + 2 = (10 - 6)/(1 - (-1))
6c + 2 = 4/2
6c + 2 = 2
6c = 0
c = 0

0 is in the interval [-1,1]

Answer 3

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