Bài tập 5.82 trang 212 SBT Toán 11

Ta có:

\(\begin{array}{l}
\Delta f\left( x \right) = f\left( {{x_0} + \Delta x} \right) - f\left( {{x_0}} \right)\\
 = {\left( {{x_0} + \Delta x} \right)^3} - 2\left( {{x_0} + \Delta x} \right) + 1 - \left( {x_0^3 - 2{x_0} + 1} \right)\\
 = x_0^3 + 3x_0^2\Delta x + 3{x_0}{\left( {\Delta x} \right)^2} + {\left( {\Delta x} \right)^3} - 2{x_0} - 2\Delta x + 1 - x_0^3 + 2{x_0} - 1\\
 = 3x_0^3\Delta x + 3{x_0}{\left( {\Delta x} \right)^2} + {\left( {\Delta x} \right)^3} - 2\Delta x\\
df\left( x \right) = \left( {3{x^2} - 2} \right)\Delta x
\end{array}\)

Suy ra: 

\(\begin{array}{l}
{\rm{\Delta }}f\left( 1 \right) = 3{\rm{\Delta }}x + 3{\left( {{\rm{\Delta }}x} \right)^2} + {\left( {{\rm{\Delta }}x} \right)^3} - 2{\rm{\Delta }}x = {\left( {{\rm{\Delta }}x} \right)^3} + 3{\left( {{\rm{\Delta }}x} \right)^2} + {\rm{\Delta }}x\\
df\left( 1 \right) = \left( {{{3.1}^2} - 2} \right){\rm{\Delta }}x = {\rm{\Delta }}x
\end{array}\)

a) Tại Δx = 1 ta có \(\left\{ \begin{array}{l}
\Delta f\left( 1 \right) = 5\\
df\left( 1 \right) = 1
\end{array} \right. \Rightarrow \Delta f\left( 1 \right) > df\left( 1 \right)\)

b) Tại \(\Delta x = 0,1 \Rightarrow \left\{ \begin{array}{l}
\Delta f\left( 1 \right) = 0,131\\
df\left( 1 \right) = 0,1
\end{array} \right. \Rightarrow \Delta f\left( 1 \right) > df\left( 1 \right)\)

c) Tại \(\Delta x = 0,01 \Rightarrow \left\{ \begin{array}{l}
\Delta f\left( 1 \right) = 0,010301\\
df\left( 1 \right) = 0,01
\end{array} \right. \Rightarrow \Delta f\left( 1 \right) > df\left( 1 \right)1\)