Question:
The approximate value of $f\left(x\right)= x^{3}+5x^{2}-7x +9$ at $x=1.1 $ is
The approximate value of $f\left(x\right)= x^{3}+5x^{2}-7x +9$ at $x=1.1 $ is
Updated On: Apr 27, 2024
- $8.6$
- $8.5$
- $8.4$
- $8.3$
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The Correct Option is A
Solution and Explanation
Given, $f(x)=x^{3}+5 x^{2}-7x+9$
On differentiating both sides w. r. t. $x$, we get
$f' (x)=3 x^{2}+10x-7$
Let $x=1$ and $\Delta x=0.1$, so that
$f(x+\Delta \,x)=f (1+0.1)=f(1.1)$
We know that,
$f(x+\Delta \,x)=f(x)+\Delta x f' (x)$
$=x^{3}+5 x^{2}-7 x+9+\Delta \,x \times \left(3 x^{2}+10 x -7\right)$
Put $x = 1$ and $\Delta x=0.1$, we get
$f (1+0.1)$
$=1^{3}+5(1)^{2}-7(1)+9+0.1 \times\left(3 \times 1^{2}+10 \times 1-7\right)$
$\Rightarrow f(1.1)=1+5-7+9+0.1(3+10-7)$
$=8+0.1(6)=8+0.6=8.6$
On differentiating both sides w. r. t. $x$, we get
$f' (x)=3 x^{2}+10x-7$
Let $x=1$ and $\Delta x=0.1$, so that
$f(x+\Delta \,x)=f (1+0.1)=f(1.1)$
We know that,
$f(x+\Delta \,x)=f(x)+\Delta x f' (x)$
$=x^{3}+5 x^{2}-7 x+9+\Delta \,x \times \left(3 x^{2}+10 x -7\right)$
Put $x = 1$ and $\Delta x=0.1$, we get
$f (1+0.1)$
$=1^{3}+5(1)^{2}-7(1)+9+0.1 \times\left(3 \times 1^{2}+10 \times 1-7\right)$
$\Rightarrow f(1.1)=1+5-7+9+0.1(3+10-7)$
$=8+0.1(6)=8+0.6=8.6$
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