1
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Sammy Black Mar 13 '13 at 6:30
  • $\begingroup$ I thought 12x is the actual derivative?? $\endgroup$ – Shane Yost Mar 13 '13 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. $\endgroup$ – Sammy Black Mar 17 '13 at 6:50
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.