Discussion Forum

 The  approximate value of $f(x) = x^{3}+5x^{2}-7x+9$  at x=1. 1 is 

Answer:

Explanation:

Given,  $f(x) = x^{3}+5x^{2}-7x+9$

 On differentiating both sides w.r.t x , we get

   f'(x) = $3x^{2}+10x-7$

Let x=1  and $\triangle x$=0.1 , so that

 $f(x+\triangle x)=f(1+0.1)=f(1.1)$

We know that  ,

 $f(x+\triangle x)=f(x)+\triangle x f'(x)$

 = $x^{3}+5x^{2}-7x+9+\triangle x\times (3x^{2}+10x-7)$

  Put x=1  and $\triangle x$=0.1 , we get

 f(1+0.1)

  = $1^{3}+5(1)^{2}-7(1)+9+0.1\times (3\times 1^{2}+10\times1-7)$

$\Rightarrow$      $f(1.1)=1+5-7+9+0.1(3+10-7)=8+0.1(6)=8+0.6=8.6$