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$f(x) = 8 b^x$

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My answer will be in terms of $b$.

I know you find the slope typically with $\dfrac{y_1-y_2}{x_1-x_2}$, but that doesn't seem to work in this situation.

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The slope of a smooth curve at a pont on the curve is defined as the slope of the tangent line at that point. Calculation of this requires some concepts from the calculus. But finding the slope of the straight line that joins $p$ and $q$ just requires the formula you quoted. Here $x_1$ and $x_2$ are given to you, but you need to calculate $y_1$ and $y_2$. –  André Nicolas Nov 27 '12 at 23:01

3 Answers 3

up vote 3 down vote accepted

Write $\,Q=(5,8\cdot b^5)\,\,,\,P=(1,8\cdot b)\,$ , then the slope between these two points is:

$$m_{PQ}=\frac{8b^5-8b}{5-1}=2b(b^4-1)$$

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So you used 8 as the y values? Interesting... it worked! –  Tyler Zika Nov 27 '12 at 23:03
    
Well, that's what the function is, not my call! If we have a function $\,y=f(x)\,$, then a point on the plane belongs to the function's graph iff the point has the form $\,(x_0,f(x_0))\,$ , for some number $\,x_0\,$ in the function's domain. This is what I applied here. –  DonAntonio Nov 27 '12 at 23:05

It looks like the original question had no calculus tag and judging by the way the question was asked, I'm guessing no knowledge of calculus.

The formula you posted for slope works for a line where slope is constant. Without knowledge of calculus, the best you can do is try to draw an approximate tangent line, the find the slope of the line.

This is a problem made for calculus though. The slope at any point on a curve $y=f(x)$ is the derivative of $y$ with respect to $x$, written as $\frac{dy}{dx}$. Conan Wong gives the correct calculation for this derivative.

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Most probably the question's title is wrongly worded: it should have been "slope of line through two points on a function's graph". –  DonAntonio Nov 27 '12 at 23:13
    
@DonAntonio It isn't quite clear from the question, but had I noticed Andre's comment, I probably would have upvoted that rather than giving this answer as it seems he had it all covered. –  Mike Nov 27 '12 at 23:23

$$b^x = e^{ln (b^x)} = e^{x ln (b)}$$

so using the Chain Rule and the fact that $(e^x)' = e^x$ we have

$$(b^x)' = (e^{xln(b)})' = (ln(b)) e^{xln(b)} = (ln (b))b^x$$

So the derivative of $8b^x$ is $8(ln(b))b^x$.

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Because of the question's wording, it was almost sure the OP didn't mean to use derivatives, which is what you did (correctly, by the way) –  DonAntonio Nov 27 '12 at 23:08
    
@DonAntonio, yeah, I only figured that out after re-reading Tyler's question. But anyway, good practice for me :-) –  Conan Wong Nov 27 '12 at 23:09

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