## University Calculus: Early Transcendentals (3rd Edition)

We are given: $f(x)=x^2+2x$ for $\delta f$ at 0.1 $\delta f=f(1+0.1)-f(1)$ or $\delta f=1.1^2+2\times1.1-1-2\times 1=1.21+2.2-1-2=0.41$ or $\delta f=0.41$ differentiation of f(x): $f'(x)=2x+2$ $df=f'(1)\times0.1$ $df=(2+2)\times0.1=0.4$ $df=0.4$ so $|\delta f-df|=|0.41-0.4|=|0.01|=0.01$ The final answer is: a) 0.41 b) 0.4 c) 0.01