Question

The approximate value of $f\left(x\right)=x^3+5x^2-7x+9$ at $x=1.1$ is

• A. $8.6$
• B. $8.5$
• C. $8.4$
• D. $8.3$

Answer 1

Answer: (A)

Given, $f(x)=x^3+5x^2-7x+9$
On differentiating both sides w. r. t. $x$, we get
$f^{\prime }(x)=3x^2+10x-7$
Let $x=1$ and $\Delta x=0.1$, so that
$f(x+\Delta x)=f(1+0.1)=f(1.1)$
We know that,
$f(x+\Delta x)=f(x)+\Delta x{f}^{\prime }(x)$
$=x^3+5x^2-7x+9+\Delta x\times \left(3x^2+10x-7\right)$
Put $x=1$ and $\Delta x=0.1$, we get
$f(1+0.1)$
$=1^3+5(1{)}^2-7(1)+9+0.1\times \left(3\times 1^2+10\times 1-7\right)$
$\Rightarrow f(1.1)=1+5-7+9+0.1(3+10-7)$
$=8+0.1(6)=8+0.6=8.6$