Answer
$a)$
$\Delta f=0.41$
$b)$
$df=0.4$
$c)$
$The\space approximation \space error =0.1$
Work Step by Step
$Given:$ $x^{2}$+$2x$ $,$ $x_{0}=1$ $and$ $dx=0.1$
$a)$
$since,$ $\Delta f=f(x_{0}+dx)-f(x_{0})$ $and$ $x_{0}+dx=1+0.1$
$\Rightarrow$ $\Delta f=[(1.1^{2})+2(1.1)]-[(1^{2})+2(1)]$
$\Rightarrow$ $\Delta f=[3.41]-[3]=0.41$
$\Rightarrow$ $\Delta f=0.41$
$b)$
$\because$ $df=\frac{df}{dx}\times{d x}$
$\frac{df}{dx}=2x+2$
$\Rightarrow$ $df= (2x_{0}+2)\times{dx}$
$\Rightarrow$ $df= (2(1)+2){0.1}$
$\Rightarrow$ $df= 0.4$
$c)$
$The\space approximation \space error=\mid \Delta f-df \mid $
$\Rightarrow$ $ \mid 0.41-0.4 \mid=0.1$
$Hence,$
$The\space approximation \space error \space is \space 0.1$
