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

Dec 29, 2016

$f \left(- 2.01\right) \approx 1.05$

#### Explanation:

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

$f \left(- 2.01\right) = f \left(- 2\right) + {\int}_{- 2}^{- 2.01} f ' \left(x\right) \text{d} x$

Using a right rectangle approximation, assume that $f ' \left(x\right) \approx f ' \left(- 2\right)$ for all $- 2.01 \le x \le - 2$

The above equation then becomes

$f \left(- 2.01\right) \approx f \left(- 2\right) + {\int}_{- 2}^{- 2.01} f ' \left(- 2\right) \text{d} x$

$= f \left(- 2\right) + \left[- 2 - \left(- 2.01\right)\right] f ' \left(- 2\right)$

$= f \left(- 2\right) + \left[- 2 - \left(- 2.01\right)\right] \left(5\right)$

$= f \left(- 2\right) + \left(0.01\right) \left(5\right)$

$= f \left(- 2\right) + 0.05$

$= 1 + 0.05$

$= 1.05$

Dec 30, 2016

$0.95$

#### Explanation:

$f \left(x + h\right) \approx f \left(x\right) + h f ' \left(x\right)$
Let $h = - 0.01$, $x = - 2$, $f \left(x\right) = 1$, $f ' \left(x\right) = 5$

$f \left(- 2 - 0.01\right) \approx \left(1\right) + \left(- 0.01\right) \times \left(5\right)$
$f \left(- 2.01\right) \approx 0.95$