The a[[rpxo,ate va,ie pf `f(x)=x^3+5x^(2)-7x+9" at "x=1` is

Options

A. 8
B. 8.5
C. 8.4
D. 8.3

Answer

A

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 `Deltax=0.1,` so that
`f(x-Deltx)=t(1+-0.1)=f(1,1)`
We know that
`f(x+Deltax)+f(x)+Deltax f'(x)`
`=x^(3)+5x^(2)-7x+9+Deltax xx (3x^(2)+10x-7)`
Put x=1 and `Delta x=0.1` we get f(1+0.1)
`=t^(3)+5(1)^(2)-7(1)+9+0.1xx(3xxt^(2)+10xx1-7)`
`Rightarrow f(1,1)=1+5-7+9+0.1(3+10-7)=8+0.1(6)=8+0.6=8.6`