12.6 Bolzano theorem
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Use Bolzano’s theorem to prove that the following equations have at least one real solution.
- \(e^x + x + 2 = 0\)
- \(x^3 + \ln(x) = -\sqrt{x}\)
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Let \(f\) be a continuous function on the closed interval \([0, 1]\) with range also contained in \([0, 1]\). Prove that \(f\) must have a fixed point; that is, show \(f(x) = x\) for at least one value of \(x \in [0, 1]\).
Suppose for all \(x\in[0,1]:f\left(x\right)\neq x\)
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\(f(x)>x\) for all \(x\in[0,1]\)
Then \(f(x)>x\geq1\) contradiction! 2. \(f(x)<x\) for all \(x\in [0,1]\)
Then \(f(x)<x\leq 0\) contradiction! 3. \(f(x)<x\) for some \(x\in [0,1]\) and \(f(x)>x\) for some \(x\in [0,1]\)
Consider \(x_1<x_2\), then \(\exists x_1:f(x_1)<x_1\) and \(\exists x_2:f(x_2)>x_2\). Consider \(g(x)=f(x)-x\)
By Bolzano theorem, since it is continuous and \(g(x_1)<0,g(x_2)>0\), then there exists \(g(x)=0\) which is \(f(x)=x\)
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Choose an interval \([a, b]\) such that the function \(f : [a, b] \to \mathbb{R}\) will be injective. Give the inverse function in each case.
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\(f(x) = x^2\)
\([0,1]\), \(y^{-1}=\sqrt{x}\) 2. \(f(x) = \frac{1}{x}\)
\([1,2]\), \(y^{-1}=x\)
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Compute the inverse of the following functions:
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\(f(x) = -2\ln\left(\frac{x+1}{x-1}\right)\)
\(f:(-\infty,-1)\cup(1,\infty)\rightarrow\mathbb{R}\)
\(y^{-1}=\frac{2}{e^{-\frac{x}{2}}-1}+1\) 2. \(f(x)=\begin{cases}\frac12x & \text{if }x<0\\ 2x & \text{if }x\geq0\end{cases}\)
\(f^{-1}(x)=\begin{cases}2x & \text{if }x<0\\ \frac12x & \text{if }x\geq0\end{cases}\) 3. \(f(x)=\begin{cases}-x^2 & \text{if }x\geq0\\ 1-x^3 & \text{if }x<0\end{cases}\)
\(f^{-1}(x)=\begin{cases}\sqrt{-x} & \text{if }x\leq0\\ \sqrt[3]{1-x} & \text{if }x>1\end{cases}\)
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Using the definition of \(\ln\), prove that:
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\(\ln(ab) = \ln(a) + \ln(b)\)
\(\ln\left(ab)=t\Rightarrow e^{t}=ab\right.\) and \(\ln a=m\Rightarrow e^{m}=a\) and \(\ln b=n\Rightarrow e^{n}=b\)
Then \(ab=e^{m}\cdot e^{n}=e^{m+n}=e^{t}\Rightarrow m+n=t\Rightarrow\ln(ab)=\ln(a)+\ln(b)\) 2. \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
\(\ln\left(\frac{a}{b})=t\Rightarrow e^{t}=\frac{a}{b}\right.\) and \(\ln a=m\Rightarrow e^{m}=a\) and \(\ln b=n\Rightarrow e^{n}=b\)
Then \(\frac{a}{b}=\frac{e^{m}}{e^{n}}=e^{m-n}=e^{t}\Rightarrow m-n=t\Rightarrow\ln(ab)=\ln(a)-\ln(b)\) 3. \(\ln(a^b) = b\ln(a)\)
\(\ln\left(a^{b})=t\Rightarrow e^{t}=a^{b}\right.\) and \(b\ln a=m\Rightarrow e^{\frac{m}{b}}=a\)
Then \(a^{b}=\left(e^{\frac{m}{b}}\right)^{b}=e^{m}=e^{t}\Rightarrow\ln(a^{b})=b\ln(a)\)
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