Estimate the slope of the tangent line to the graph of a reciprocal function
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mmm4444bot
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Have you ever used a graphing calculator to zoom in on a particular point of a graph? When you zoom in real close (that is, when you look at the curve over a tiny interval of x values), the curve appears to be a straight line. The slope of that line is very close to the slope of the tangent line.
One way to estimate the slope of the tangent line is to use the slope formula with values that you calculate.
For this exercise, pick any two x values that are very close to 3 -- one less than 3 and the other greater than 3.
For example, you could pick x1=2.99 and x2=3.01
Calculate the corresponding values for y1 and y2
Use the slope formula with your values for (x1,y1) and (x2,y2)
Questions about this? Work to show?
One way to estimate the slope of the tangent line is to use the slope formula with values that you calculate.
For this exercise, pick any two x values that are very close to 3 -- one less than 3 and the other greater than 3.
For example, you could pick x1=2.99 and x2=3.01
Calculate the corresponding values for y1 and y2
Use the slope formula with your values for (x1,y1) and (x2,y2)
Questions about this? Work to show?
HallsofIvy
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I'm confused. If you are asked to find "the slope of the tangent line", are you not taking a Caclulus course? If you are then you should certainly know that the "slope of the tangent line" at a given value of x is the derivative of the function there and the quotient rule. What do you get when you differentiate \(\displaystyle \frac{x}{x- 2}\) using the quotient rule?
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mmm4444bot
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JSmith, do you see what's happening, in this thread. Everybody is guessing at what you've been asked to do.
Please re-read the main points from our posting guidelines. Then, come back here and post something about the course from which these estimation exercises come AND what your class has been talking about lately.
Doing this may help prevent guessing games. Cheers :cool:
Please re-read the main points from our posting guidelines. Then, come back here and post something about the course from which these estimation exercises come AND what your class has been talking about lately.
Doing this may help prevent guessing games. Cheers :cool:
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If you compute the difference quotient
\(\displaystyle \dfrac{f(x+h)-f(x)}{h}=\dfrac{\dfrac{x+h}{x+h-2}-\frac{x}{x-2}}{h}=\dfrac{(x+h)(x-2)-x(x+h-2)}{h(x+h-2)(x-2)}=\)
\(\displaystyle \dfrac{x^2-2x+hx-2h-x^2-hx+2x}{h(x+h-2)(x-2)}=\dfrac{-2h}{h(x+h-2)(x-2)}=-\dfrac{2}{(x+h-2)(x-2)}\)
Now as h gets smaller and smaller, i.e., \(\displaystyle h\to0\) this becomes:
\(\displaystyle -\dfrac{2}{(x-2)^2}\)
And when \(\displaystyle x=3\), the slope is:
\(\displaystyle -\dfrac{2}{(3-2)^2}=-2\)
\(\displaystyle \dfrac{f(x+h)-f(x)}{h}=\dfrac{\dfrac{x+h}{x+h-2}-\frac{x}{x-2}}{h}=\dfrac{(x+h)(x-2)-x(x+h-2)}{h(x+h-2)(x-2)}=\)
\(\displaystyle \dfrac{x^2-2x+hx-2h-x^2-hx+2x}{h(x+h-2)(x-2)}=\dfrac{-2h}{h(x+h-2)(x-2)}=-\dfrac{2}{(x+h-2)(x-2)}\)
Now as h gets smaller and smaller, i.e., \(\displaystyle h\to0\) this becomes:
\(\displaystyle -\dfrac{2}{(x-2)^2}\)
And when \(\displaystyle x=3\), the slope is:
\(\displaystyle -\dfrac{2}{(3-2)^2}=-2\)
mmm4444bot
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How does this look:
Your calculations look good, but you did not correctly substitute your x and y values into the slope formula.
The difference of the y-coordinates goes on top, and the difference between the x-coordinates goes on bottom.
In other words, you determined the reciprocal of the estimated slope at x=3.
The estimated slope of the tangent line is -2.