- #1
sp09ta
- 10
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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.
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.