Find the approximate value of `f(2. 01)`, where `f(x)=4x^2+5x+2`.

Explanation

Here, we will use the definition of derivatives:

`f'(x) = (f(x+Deltax)-f(x))/((x+Deltax)-x)`

`f'(x) = (f(x+Deltax)-f(x))/(Deltax)`

Here, `x = 2 and Deltax = 0.01`

`:. f'(2) = (f(2+0.01) - f(2))/0.01->(1)`

Here, `f'(x) = 8x+5`

`:. f'(2) = 8(2)+5 = 21`

`f(2) = 4(4)+5(2)+2 = 28 `

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