Mean Value

Question

Determine if the Mean Value Theorem for Integrals applies to the function f(x) = 3 - x^2 on the interval ⌈0, √3⌉. If so, find the x-coordinates of the point(s) guaranteed by the theorem.

No, the Mean Value Theorem for Integrals does not apply

Yes, x = 1

Yes, x = 2

Yes, x= plus or minus 1

Answer 1

In order for the mean value theorem to be applied, the given function must be continuous and differentiable at the given intervals.  The given function is a parabola, so the theorem applies here.

 

According to the mean value theorem, the slope between two points must be the same as the derivative of the function at any point c.

 

In other words,

 

[f(b) - f(a)] / (b - a) = f'(c)

 

where:

a = 0

b = √3

 

a < c < b

 

Note that the slope is the same as the derivative.  The derivative is the slope of the tangent line.

 

 

Now we find the slope between the two points in the interval.

 

[3 - √(3)² - 3] / (√(3) - 0)) =

 

-3 / √3

 

 

Now we set the derivative of f(c) equal to -3/√3 and solve for c

 

f'(c) = -3 / √3

 

-2c = -3 / √3

 

 

c = (-1 / 2)(-3 / √3)

 

c = 3 / (2√3)

 

c = 0.866

 

 

a < c < b

 

0 < 0.866 < 1.732

 

 

This verifies the theorem.  x=c.   So   x=1

 

Answer 2

The function is a polynomial and hence continuous on the given interval. Therefore you can use the Mean Value Theorem as follows:

(b-a)f(c) = ∫^ba f(x)dx

(√3)(3-c²)=[3x-x³/3]|₀^√3

(√3)(3-c²)=3√3-(√3)³/3

3-c²=3-1

3-c²=2

c²=1

c=±1

so, the answer is Yes, x=1  since as Michael J. reminded me  that -1 is not in the given interval!