The approximate value of the function $\mathrm{f}(x)=x^{3}+5 x^{2}-7 x+10$ at $x=1 \cdot 1$ is
The approximate value of the function $\mathrm{f}(x)=x^{3}+5 x^{2}-7 x+10$ at $x=1 \cdot 1$ is
$7 \cdot 6$
$8 \cdot 6$
$6 \cdot 6$
$9 \cdot 6$
Solution
Given $f(x)=x^{3}+5 x^{2}-7 x+10 \Rightarrow f^{\prime}(x)=3 x^{2}+10 x-7$
Let $a=1, h=0.1$
$\therefore \mathrm{f}(1)=1+5-7+10=9$ and $\mathrm{f}^{\prime}(1)=3+10-7=6$
We know that $\mathrm{f}(\mathrm{a}+\mathrm{h}) \div \mathrm{f}(\mathrm{a})+\mathrm{hf}^{\prime}(\mathrm{a})$
$f(1.1)=9+(0.1)(6)=9+0.6=9.6$