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# Let$\Delta f = f ( 1 + h ) - f ( 1 ),$where$f ( x ) = x ^ { - 1 }.$Show directly that$E = \left| \Delta f - f ^ { \prime } ( 1 ) h \right|$is equal to$h ^ { 2 } / ( 1 + h ).$Then prove that$E \leq 2 h ^ { 2 }$if$- \frac { 1 } { 2 } \leq h \leq \frac { 1 } { 2 }.$Hint: In this case,$\frac { 1 } { 2 } \leq 1 + h \leq \frac { 3 } { 2 }.$

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$\begin{array}{l} Need\,\,to\,\,prove\,\,\\ \\ E \le 2{h^2}\\ \\ f(x) = {x^{ - 1}}\\ \\ f(x) = \frac{1}{x}\\ \\ \Delta f = f(1 + h) - f(1)\\ \\ = \frac{1}{{1 + h}} - 1\\ \\ = \frac{{1 - 1 - h}}{{1 + h}} = - \frac{h}{{1 + h}} \end{array}$

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