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Find the slope of the secant to the curve f(x)=-3logx+2 between these points:

Vela

New member
Jan 9, 2014
2
The question was too long to post in the title so I just wrote down the first part. I hope this is alright. Here is the question that I am doing right now:



This is the graphical representation (thanks to Desmos Graphing Calculator):



So I have substituted the points in the equation to get their respective y-values.
For example:

f(1.1) = -3log(1.1)+2
f(1.1) =~ 1.87

I've done the questions myself using this method to find the slope of the secant line, and I wanted confirmation that I was doing it right.

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I've written it on paper and unfortunately I don't have a scanner so I will just type out (i) and (ii) to show my rationale.

4.a)

i] m = (Y2-Y1) / (X2-X1)
= (1.097-2) / (2-1)
= -0.903 / 1
m = -0.903

ii] m = (Y2-Y1) / (X2-X1)
= (1.472-2) / (1.5-1)
= -0.528 / 0.5
m = -1.056

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I just wanted to know if I was doing it correctly, and if not how can I answer this question? Thanks in advance.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Your example shows you're using $\log(x)=\log_{10}(x)$. If so, it appears your first one is correct; I didn't bother to check the second, as it's analogous. Looks fine to me!
 

Vela

New member
Jan 9, 2014
2
OK, thank you. I just have one more question:



This is the second part to question #4. I wanted to do it myself before I confirmed my answer here.

Here are my steps:

I found the derivative of the initial equation: f(x) = -3log(x)+2

The derivative was: f(x) = (-3) / (ln[10] x (x))

So I substituted 1 for x, and the answer I got was -1.303. I rounded it to 3 decimal places. Is this the correct method, or is there another way?

Thanks in advance.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, you are doing fine. I think what I would do is recognize all four questions are the same except for one parameter, $\Delta x$. So I would derive a formula into which I could then just plug into. In other words, work the problem once instead of four times.

The slope is given by:

\(\displaystyle m=\frac{\Delta f}{\Delta x}=\frac{f\left(x+\Delta x \right)-f(x)}{\Delta x}\)

Now, using the given definition of $f(x)$, we may write:

\(\displaystyle m=\frac{\left(-3\log\left(x+\Delta x \right)+2 \right)-\left(-3\log\left(x \right)+2 \right)}{\Delta x}=\frac{3\log\left(\frac{x}{x+\Delta x} \right)}{\Delta x}\)

With $x=1$, there results:

\(\displaystyle m=\frac{3\log\left(\frac{1}{1+\Delta x} \right)}{\Delta x}\)

a) \(\displaystyle \Delta x=1\)

\(\displaystyle m=\frac{3\log\left(\frac{1}{1+1} \right)}{1}=-3\log(2)\approx-0.903\)

b) \(\displaystyle \Delta x=\frac{1}{2}\)

\(\displaystyle m=\frac{3\log\left(\frac{1}{1+\frac{1}{2}} \right)}{\frac{1}{2}}=6\log\left(\frac{2}{3} \right)\approx-1.057\)

Now just plug-n-chug for the remaining two. :D

The formula we derived will be very useful in answering part b). Let $\Delta x$ get smaller and smaller until two successive results agree to 3 decimal places. Or, as you did, you can use calculus, but I suspect you are not expected to be able to differentiate.