1) Find the value “c” that shows the mean value theorem for derivatives holds for f(x𝑥) =2x^𝑥2 on the interval [−3,1]

1) Find the value “c” that shows the mean value theorem for derivatives holds for f(x) =2x^2 on the interval [−3,1] 

The slope of the line drawn between f(-3)  and f(1) =     [ 2 - 18] / [ 1 - (-3) ] =   -16/ 4 = -4

And the derivative of 2x^2  =  4x

So.....we're looking for the x value where the tangent line to the curve at that point has the slope of -4

4x = -4     divide through by 4

x = -1

And    -3 < -1 < 1       so the Theorem holds

2) Find the value “c” that shows the mean value theorem for integrals holds for 𝑓f(x) =2x^2 on the interval [1,6] 

We are trying to find "c"  such that

6

∫   2x^2 dx =   (2)c^2 * (6 -1)

1

 (2/3)[6^3 - 1^3]  = 10c^2

(2/3) (215)  = 10c^2

215/15 = c^2

43/3  = c^2   take the positive root

√[43/3]  = c ≈  3.7859

And this number is in the interval [1,6]