#### Answer

a) 0.02
b) 0
c) 0.02

#### Work Step by Step

We are given: $f(x)=2x^2+4x-3$
for $\delta f$ at 0.1:
$\delta f=f(-1+0.1)-f(-1)$
or $\delta f=2(-0.9)^2+4\times0.9-3-2\times(- 1)+4(-1)-3=-4.98-(-5)=0.02$
or$\delta f=0.02$
differentiation of f(x):
$f'(x)=4x+4$
$df=f'(-1)\times0.1$
$df=(-4+4)\times0.1=0.0$
$df=0.0$
so
$|\delta f-df|=|0.02-0.0|=|0.02|=0.02$
The final answer is: a)0.02, b)0, c)0.02