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

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

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Answer 1 · 2016-12-29T11:05:37

$f (- 2.01) \approx 1.05$ $f(-2.01) ~~ 1.05$

Explanation:

Assuming that $f (x)$ $f(x)$ is differentiable in the neighborhood of $x = 2$ $x=2$ , 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 · 2016-12-30T08:29:39

$0.95$

Explanation:

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

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

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

2 Answers

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