Answer :
[tex]f(1.2)\approx f(1)+f'(1)(1.2-1)=4+3\times0.2=4.6[/tex]
Compare to the actual value:
[tex]\begin{cases}f'(x)=2x+1\\f(1)=4\end{cases}\implies f(x)=x^2+x+2\implies f(1.2)=4.64[/tex]
Compare to the actual value:
[tex]\begin{cases}f'(x)=2x+1\\f(1)=4\end{cases}\implies f(x)=x^2+x+2\implies f(1.2)=4.64[/tex]