Help ASAP (Average Slope)

[thomcart8]

For f(x) = ln(x), estimate f'(2) by finding the average slope over intervals that get smaller and smaller, but still contain the value x = 2.

Ok so far I have gotten .1 = .487902, .01 = .498754, .001 = .499875.

Am i even doing this right?

I plugged (ln(2+.1)-(2))/.1 to get my first answer. Can someone tell me if i am even on the right track?

 

[galactus]

The slope of ln(x) at x=2 is 1/2.

What are you appraoching in your attempts?.

Does it look like you're getting closer and closer to 1/2?.

 

[thomcart8]

actually because of what i am seeing i would say that i am getting closer to 1/2

 

[thomcart8]

plus i dont know if anyone can see this but i am trying to estimate for f ' (2)

 

[galactus]

The average rate of change of y w.r.t x over an interval is the slope, m, of the secant line

joining the points.

\(\displaystyle m=\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}\)

So, try some points and hone in on the slope at x=2.

Try:

\(\displaystyle \frac{ln(2.5)-ln(1.5)}{2.5-1.5}=.5108\)

Now, maybe try:

\(\displaystyle \frac{ln(2.1)-ln(1.9)}{2.1-1.9}=.5004\)

See?. Is that what you mean?.

If so, get closer by trying 2.01 and 1.99 and so on.

As \(\displaystyle x_{1}\to x_{0}\) then we end up with the slope at a particular point.

Hence, we get the definition of a derivative or the instantaneous rate of change of y w.r.t x at the point \(\displaystyle x_{0}\)

\(\displaystyle \lim_{x_{1}\to x_{0}}\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}\)

i.e. the closer \(\displaystyle x_{1}\) gets to \(\displaystyle x_{0}\), the closer we get to the point where we want our slope.