Mean Value Theorem - Find the values of c

[sp09ta]

1. Verify that the function satisfies the mean value theorem, then find all numbers c that satisfy the the conclusion of the mvt

f(x)=e^(-2x) on the interval [0,3]

2. f'(c)=[f(b)-f(a)]/[b-a]
3.
1. f(x) is a composition of continuous functions, so f(x) is continuous on[0,3].
2. f(x) is a composition of differentiable functions, so it is differentiable on (0,3).

so, f(b)-f(a)=f'(c)(b-a)
e^-6-e^0=(-2*e^-2c)(3)

Then take the ln of both sides...
-6-0=ln(-6*e^-2c)
-6=ln(-6)+2c

Solve for c..?
c=[-6-ln(-6)]/[-2]

Can't have ln of a negative..(well without imaginary numbers). What's gone wrong? I'm stumped.

 

[LCKurtz]

You have
e^-6 - e⁰=(-2*e^-2c)(3) = -6e^-2c

At that point both sides of the equation are negative. So multiply both sides by -1 before taking logarithms.

 

[HallsofIvy]

1. Verify that the function satisfies the mean value theorem, then find all numbers c that satisfy the the conclusion of the mvt

f(x)=e^(-2x) on the interval [0,3]

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



3.
1. f(x) is a composition of continuous functions, so f(x) is continuous on[0,3].
2. f(x) is a composition of differentiable functions, so it is differentiable on (0,3).

so, f(b)-f(a)=f'(c)(b-a)
e^-6-e^0=(-2*e^-2c)(3)

Then take the ln of both sides...
-6-0=ln(-6*e^-2c)
-6=ln(-6)+2c

NO. ln(e^{-6}- e^0) is NOT -6-0. In general, ln(a+b) is NOT ln(a)+ ln(b).

Solve for c..?
c=[-6-ln(-6)]/[-2]

Can't have ln of a negative..(well without imaginary numbers). What's gone wrong? I'm stumped.

 

[sp09ta]

Hehehe... I had my value for b-a incorrect, which threw off my whole solution.. :D Got the right answer now, thanks for the feedback ;)