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Exercise 7.3 · Q10

Q.Using mean value theorem prove that for, a>0,b>0,∣e−a−e−b∣<∣a−b∣a>0,\\ b>0,\\ |e^{-a}-e^{-b}|<|a-b|.

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Apply LMVT to f(x)=e−xf(x)=e^{-x} on the interval between aa and bb; the resulting factor e−ce^{-c} is strictly between 00 and 11 because c>0c>0, which turns the LMVT equality into the required strict inequality.

Step 1. Set up LMVT for f(x)=e−xf(x)=e^{-x}.

Assume, without loss of generality, a<ba<b (both >0>0). f(x)=e−xf(x)=e^{-x} is continuous and differentiable everywhere, so LMVT on [a,b][a,b] gives some c∈(a,b)c\in(a,b) with

f′(c)=f(b)−f(a)b−a ⇒ −e−c=e−b−e−ab−a ⇒ e−b−e−a=−e−c(b−a).f'(c)=\frac{f(b)-f(a)}{b-a}\ \Rightarrow\ -e^{-c}=\frac{e^{-b}-e^{-a}}{b-a}\ \Rightarrow\ e^{-b}-e^{-a}=-e^{-c}(b-a).

Step 2. Take absolute values.

∣e−a−e−b∣=∣−e−c(b−a)∣=e−c ∣b−a∣=e−c ∣a−b∣|e^{-a}-e^{-b}|=|{-e^{-c}(b-a)}|=e^{-c}\,|b-a|=e^{-c}\,|a-b|

(using e−c>0e^{-c}>0 always, and ∣b−a∣=∣a−b∣|b-a|=|a-b|).

Step 3. Bound the factor e−ce^{-c}. …

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