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.
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