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I have missed something in relation to finding the line of tangency with the curve. I thought the line of tangency should touch the point of interest but NOT intersect when we are finding the instantaneous rate of change of that point of interest.

I wanted to include pictures but I do not have enough reputation points to do so. Could someone bump me up so I can develop more clear and concise questions in the future.

I'm trying to attain the function of the tangent line to a curve at a given point for finding the instantaneous rate of change. I know both functions of the curve and the tangent line will have the common x value. Therefore when I find the derivative I will plug the x value (x value of the point of interest) in the derivative and say "h" or "delta x" is approaching zero. Then I will solve leaving me with the y value of the tangent line. Therefore, that y value over the x value will be the rate of change. I believe I have the right idea here and if I don't please correct me.

However, I'm finding different functions of the derivative. This is where I would post some pics but I can't. So I'll type them below. Also every function below that I found does not touch the curve. I thought the tangent line will touch the curve but none of them do. I started plugging stuff into my graphing simulator and found that 7x will create a line that touches the curve but NOT at the point (1,8).

THE FUNCTION OF THE CURVE IS:

f(x) = 6x^2 + 2

and the point of interest is (1,8)

BY USING THE POWER RULE I FOUND THE FOLLOWING DERIVATIVE:

f'(x) = 12x

BY USING THE FORMULA OF THE LIMIT I FOUND THE FOLLOWING DERIVATIVE:

f'(x) = 12x + 6h

BY USING THE PRECEDING DERIVATIVE AND SAYING AS "h" APPROACHES 0 I FOUND THE FOLLOWING:

f'(x) = 12x

BY USING THE SAME DERIVATIVE 12x + 6h AND SAYING AS "x" APPROACHES 0 I FOUND THE FOLLOWING:

f'(x) = 6x

None of the above functions create a line that will touch the curve at the point of interest which might I add is (1,8). After plugging and playing around I found that the function 7x will in fact touch the curve but NOT at the point of interest.Therefore, what am I confusing for the derivative and what do the other functions I found mean? Also where is the actual tangent line that touches the curve at the point (1,8)?

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  • $\begingroup$ When you write f'(x) = 6x, you're describing the derivative (the slope of the tangent line) at x. At the particular point (1, 8), you have x = 1, so the slope is f'(1) = 6(1) = 6. In other words, the tangent line has slope 6 and passes through that point. Can you write down the equation of such a line? $\endgroup$ Mar 13, 2013 at 6:30
  • $\begingroup$ I thought 12x is the actual derivative?? $\endgroup$
    – Shane Yost
    Mar 13, 2013 at 18:04
  • $\begingroup$ You're right. Sorry. The derivative of the function $f(x) = 6x^2 +2$ is $f'(x) = 12x$, as you say. I was using your last formula (which is incorrect) to make a broader point. The derivative function gives the slopes of all tangent lines. You have to evaluate the function at a particular choice of $x$ in order to find the slope of a particular tangent line. $\endgroup$ Mar 17, 2013 at 6:50

1 Answer 1

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Equation of tangent at point $(x_0, f(x_0))$ is $$y = f(x_0) + f'(x_0) \cdot (x-x_0)$$

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