Mean value theorem

Question

f(x)=e^−6x,[0,4]?

find c

Answer 1

Given function f(x) = e^-6x

Given interval [ 0 , 4 ]

a = 0 , b = 4

f'(x) = -6e^-6x

By using mean value theorem,if f(x) is continuous on the interval [ 0 , 4 ] ,  f(x) is differentiable on ( 0 , 4 )

then there is at least one number c in the interval ( 0 , 4 ) (that is 0 < c < 4)

such that

f(a) = f(0) = e^-6*0 = 1

f(b) = f(4) = e^-6*4 = e^-24

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

f'(c) = [ f(4) - f(0) ] / ( 4 - 0 )

f'(c) = [ e^-24 - 1 ] / 4

-6e^-6c = [e^-24- 1] / 4

(6*4)e^-6c = -[e^-24- 1]

24e^-6c = [1- e^-24]

Apply log each side with base e.

ln (24e^-6c ) = ln (1 - e^-24)

Apply formula : ln (ab) = ln a + ln b

ln 24 + log e^-6c  = ln (1- e^-24)

Apply formula : lnb^a = a lnb  and Substitute e^-24 = 0 , ln 24 = 3.178.

3.178 + -6c lne  = ln (1)

Substitute  ln 1 = 0 , ln e = 1

3.178 + -6c*1   = 0

3.178 = 6c

c = 3.178/6

c = 0.53

The solution for f(x) = e^-6x using mean value theorem is  c = 0.53.