Question

If f(-2)=1 and f'(-2)=5, what is the approximate value of f(-2.01)?

Answer 1

`f(-2.01) ~~ 1.05`

Explanation:

Assuming that `f(x)` is differentiable in the neighborhood of `x=2`, from the Second Fundamental Theorem of Calculus,

`f(-2.01) = f(-2) + int_{-2}^{-2.01} f'(x) "d"x`

Using a right rectangle approximation, assume that `f'(x) ~~ f'(-2)` for all `-2.01 <= x <= -2`

The above equation then becomes

`f(-2.01) ~~ f(-2) + int_{-2}^{-2.01} f'(-2) "d"x`

`= f(-2) + [-2-(-2.01)] f'(-2)`

`= f(-2) + [-2-(-2.01)] (5)`

`= f(-2) + (0.01) (5)`

`= f(-2) + 0.05`

`= 1 + 0.05`

`= 1.05`

Answer 2

`0.95`

Explanation:

`f(x+h)approx f(x)+hf'(x)`
Let `h=-0.01`, `x=-2`, `f(x)=1`, `f'(x)=5`

`f(-2-0.01)approx (1)+(-0.01)xx(5)`
`f(-2.01)approx 0.95`