MEAN VALUE THEOREM

PROBLEM: 

Using Mean Value Theorem prove the inequality:   `0<\frac1x\log\left(\frac{e^x-1}x\right)<1`


SOLUTION:

Let f(x) = `e^x`. f(x) is continuous in [0, x] and differentiable in (0, x). So, by Langrange's Mean Value Theorem we can say that `\exists` at least one value of c in (0, x) such that 

f '(c) = `\frac{f(x)-f(0)}{x-0}`  `\Rightarrow` `e^c` = `\frac{e^x-1}{x}`


Now, 0 < c < x `\Rightarrow` 1 < `e^c` < `e^x` 

`\Rightarrow\1\<\frac{e^x-1}x\<\e^x\Rightarrow\0\<\log(\frac{e^x-1}x)\<\x`
`\Rightarrow` `0<\frac1x\log\left(\frac{e^x-1}x\right)<1`

Q.E.D.

Comments

Popular posts from this blog

Mathematical Artifacts: Exploring Ancient Mathematical Discoveries