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Chapter 5: Problem 40

Use Lagrange's theorem to prove that \(1+x0\).

Short Answer

Expert verified
Using Lagrange's theorem, we first define the function \(f(x) = e^x\) and apply the theorem on intervals \([0, x]\) and \([-x, 0]\), with \(x > 0\). For the interval \([0, x]\), we find there exists a \(c_1\) such that \(1+x< e^x\). For the interval \([-x, 0]\), we find there exists a \(c_2\) such that \(e^x<1+xe^x\). Therefore, the inequality \(1+x0\) holds.

Step by step solution

01

Define the Functions

Define \(f(x) = e^x\). Notice that \(f'(x) = e^x\), which means the function is increasing. We will use Lagrange's theorem for this function and different intervals to prove the inequality.
02

Apply Lagrange's theorem on the interval \([0, x]\) to find the lower bound of \(e^x\)

On the interval \([0, x]\), with \(x > 0\), Lagrange's theorem states there exists a \(c_1\) in this interval such that \(f'(c_1)=\frac{f(x)-f(0)}{x-0}\). Here, \(f'(x) = e^x\), so \(f'(c_1) = e^{c_1}\). Moreover, \(f(x) = e^x\) and \(f(0) = e^0 = 1\). Therefore, the equation becomes \(e^{c_1} = \frac{e^x - 1}{x}\). Since \(c_1\) is in the interval \((0, x)\), we have \(0 < c_1 < x\), and as \(e^x\) is an increasing function, it follows that \(1 < e^{c_1} < e^x\). Then, \(1+x< e^x\).
03

Apply Lagrange's theorem on the interval \([-x, 0]\) to find the upper bound of \(e^x\)

On the interval \([-x, 0]\), with \(x > 0\), Lagrange's theorem states there exists a \(c_2\) in this interval such that \(f'(c_2)=\frac{f(0)-f(-x)}{0 - (-x)}\), so \(f'(c_2) = e^{c_2}\). Here, \(f(x) = e^x\) and \(f(-x) = e^{-x}\). Therefore, the equation becomes \(e^{c_2} = \frac{1 - e^{-x}}{x}\). Since \(c_2\) is in the interval \((-x, 0)\), we have \(-x < c_2 < 0\), and as \(e^x\) is an increasing function, it follows that \(e^{-x} < e^{c_2} < 1\). Then, we have: \[e^x \left(\frac{1-e^{-x}}{x} \right)=1+xe^{c_2}(1-e^{-x})<1+xe^x(1-e^{-x})\] Thus, \(1 +x0\). In conclusion, using Lagrange's theorem, we have proved the inequality \(1+x0\).

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