Cho hàm số \(f\left( x \right) = {x^3} - 2x + 1\)
Hãy tính \({\rm{\Delta }}f\left( 1 \right),df\left( 1 \right)\) và so sánh chúng, nếu:
a) \({{\rm{\Delta }}x = 1}\)
b) \({{\rm{\Delta }}x = 0,1}\)
c) \({{\rm{\Delta }}x = 0,01}\)
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
\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 \(\Delta x = 0,01 \Rightarrow \left\{ \begin{array}{l}
\Delta f\left( 1 \right) = 0,010301\\
df\left( 1 \right) = 0,01
\end{array} \right. \Rightarrow \Delta f\left( 1 \right) > df\left( 1 \right)1\)
-- Mod Toán 11