Mean Value Theorem Problem

[Jason76]

\(\displaystyle f(x) = \ln x\)

\(\displaystyle [1,4]\)

\(\displaystyle \dfrac{f(4) - f(1)}{4 - 1}\)

\(\displaystyle \dfrac{1.386294361 - 0}{3}\)

\(\displaystyle \dfrac{1.386294361}{3} = 0.46209812\)

\(\displaystyle f'(x) = \dfrac{1}{x}\)

\(\displaystyle \dfrac{1}{x} = 0.46209812\)

\(\displaystyle x = \dfrac{1}{0.46209812}\)

\(\displaystyle x = 2.164042563\) Computer says this isn't right.

 

[stapel]

\(\displaystyle x = 2.164042563\) Computer says this isn't right.

Since you haven't told us what question you're supposed to be answering, nor in what format, it's impossible to say whether you're close or not. It really would be helpful if you'd start including that information, rather than continuing to require us to guess.

 

[Jason76]

Since you haven't told us what question you're supposed to be answering, nor in what format, it's impossible to say whether you're close or not. It really would be helpful if you'd start including that information, rather than continuing to require us to guess.

Iv'e basically done the whole problem. But something somewhere is wrong, cause the computer won't accept the answer.

 

[pka]

Iv'e basically done the whole problem. But something somewhere is wrong, cause the computer won't accept the answer.

You really should post the complete question.
From the title of the post I guess that you are to find a
\(\displaystyle c\in [1,4]\) such that \(\displaystyle f'(c)=\dfrac{1}{c}=\dfrac{f(4)-f(1)}{4-1}\).

 

[Jason76]

The computer didn't want decimal form. That was the problem. But the procedure was correct. The answer (by avoiding decimals) came out to be \(\displaystyle \dfrac{3}{\ln 4}\)

 

[srmichael]

The computer didn't want decimal form.

This will pretty much always be the case, especially when the number is irrational.