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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?

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  • $\begingroup$ It's just the line that passes through those two points on the graph. $\endgroup$
    – Michael Hardy
    Feb 14, 2015 at 23:16
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    $\begingroup$ Draw the line joining $(1,\ln(1))$ to $(6,\ln(6))$. This is often called a secant line. What you have computed is the slope of that line. $\endgroup$
    – André Nicolas
    Feb 14, 2015 at 23:16
  • $\begingroup$ @AndréNicolas Is this the slope of the tangent line? $\endgroup$
    – c0der
    Feb 14, 2015 at 23:43
  • $\begingroup$ It is not the slope of the tangent line, at least, not the tangent line at $x=1$, nor the tangent line at $x=6$. The slope of the tangent line at a point $x=a$ will be the limit as $h\to 0$ of slopes of secant lines that go through $(a,\ln a)$ and $(a+h,\ln(a+h))$. That is coming up very soon in your course. $\endgroup$
    – André Nicolas
    Feb 14, 2015 at 23:54

2 Answers 2

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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.

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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 $1 < c < 6.$

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