Average rate of change of $\ln(x)$, decimal slope?

Question

First of all, I'm new to calculus, so please excuse me if my question sounds silly.

I have a graph of the function $f(x)= \ln(x)$, and I want to compute the average rate of change of f on the interval $[1, 6]$, and then draw a straight line representing its slope.

The problem is, when I try to find the average rate of change, I get a decimal slope, and don't know know to add this slope line to the graph, or where to put it.

These were my steps to finding the average rate of change:

$$\frac{f(6) - f(1)} 5 = \frac{\ln(6) - \ln(1)} 5 = \frac{\ln(5)}5 = 0.32188$$

How do I finish this problem, and where do I draw this slope line?

Answer 1

First of all $\ln 6 - \ln 1 = \ln 6 - 0 = \ln 6$. So the average rate of change is given by $$\dfrac{\ln 6 - \ln 1}{6 - 1} = \dfrac {\ln 6}{5}\approx 0.358.$$

Now, use the points $(1, \ln 1) = (1, 0)$ and $(6,\ln 6)$ to draw your line (the line that passes through those two points). The average rate of change on the interval is the slope of this line.

Answer 2

what you are asked to do is to draw the secant line connecting $(1,0)$ and $(6,\ln 6)$ on the graph $y = \ln x.$ then you are invited to verify/appreciate the average rate you computed, yes it is a decimal number, as the slope of this secant line. later in calculus you will lear about the mean value theorem where it will be shown that this slope matches the slope of the tangent line to the graph $y = \ln x$ at some point $(c, \ln c)$ for $1 < c < 6.$