Help with an AP Calculus problem please

Question

amistre64 · Answer

have you tried typing in the problem?

3psilon · Answer

attachment

3psilon · Answer

@amistre64  I have attached a file . Would you be kind enough to help?

amistre64 · Answer

well, we know what the value of 5^1 is  and we have an approximation to 5^(1.002)

f' is approximately the slope between 2 points that are very close together

amistre64 · Answer

what would you recall as the slope formula from algebra when given 2 points?

3psilon · Answer

y2-y1/x2-x1

amistre64 · Answer

good

use (1,5)  and (1.002,5.016)

3psilon · Answer

But why?

anonymous · Answer

$f'(1)\approx\frac{5^{1.002}-5^1}{1.002-1}\approx\frac{5.016-5}{0.002}=\frac{0.016}{0.002}=8$

amistre64 · Answer

i thought i covered that .... 

f' defines the slope of a line at a single point.  we can approximate that slope by using to points that are very close together

3psilon · Answer

ok my bad....

anonymous · Answer

@3psilon  we are approximating the instantaneous rate of change at $x=1$ by using a secant line approximation! Recall this is just the *average* rate of change between $x=1$ and $x=1.002$, points that are rather close to one another.

amistre64 · Answer

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