Find the approximate value of `f(3.01)` when
y = `f(x) = x^(2)+3x+1`.

Explanation

`"Let "x=3 and x+Deltax=3.01`
` rArr Deltax= 0.01`
`"At "x=3, y=f(3)=3^(2)+3(3)+1=19`
`"Now " y=x^(2)+3x+1`
` rArr (dy)/(dx) = f'(x)=2x+3`
`"At "3,`
` (dy)/(dx)=f'(x)=2(3)+3=9`
`:. Deltay=(dy)/(dx).Deltax=9(0.01)=0.09`
`and y+Deltay=19+0.09`
` rArr f(3.01) = 19.09`.