I get two different values when calculating the tangent of a curve.

In summary, the slope of the tangent line to the curve y=x-x^3 at point (1,0) is 2, using both the 1st and 2nd definition of a limit. However, there were errors in the substitution process in the second equation, which led to the incorrect value of -2.
  • #1
g4w4gg
2
0
1. (a) Find the slope of the tangent line to the curve y=x-x^3 at point (1,0)
(i) using the 1st definition of a limit: lim(x->a)- (f(x)-f(a))/(x-a)
(ii) using the 2nd equation of a limit: lim(h->a)- (f(a+h)-f(a))/h​

The Attempt at a Solution


In my attempt I got two different values (the same integer value, except one's a negative).

i) (0-(x-x^3))/(x-1)= (-x+x^3)/(x-1)

Then I factored an x out of the numerator which gives me a difference of squares.

(x(x^2-1))/(x-1) = (x(x-1)(x+1))/(x-1)

Then (x-1) cancels out

=x(x+1)

Plug in lim(x->1)

=1(1+1)=2

Then I tried the other equatin:

lim(h->0)- (f(a+h)-f(a))/h

since 1 is the limit, we put 1 in as the a and it equals:

((1+h-(h+3)^3)-(1-1^3))/h

and 1 minus 1 cubed=0 so we drop the minus sign at the end along with the zero and get:

=(1+h-(h^3+3h^2+3h+1))/h

=(1+h-h^3-3h^2-3h-1)/h

=(-h^3-3h^2-2h)/h

=(-h^2-3h-2)

Then we plug in the lim(h->0)

=(-0^2-3(0)-2

lim(h->0)=-2

So as you can see, by doing both equations I get 2 and -2, but only one is tangent to the curve and the other one crosses the graph but never touches it again. I re-did these problems three times to check my calculations and nothing seems off.
 
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  • #2
g4w4gg said:
1. (a) Find the slope of the tangent line to the curve y=x-x^3 at point (1,0)
(i) using the 1st definition of a limit: lim(x->a)- (f(x)-f(a))/(x-a)
(ii) using the 2nd equation of a limit: lim(h->a)- (f(a+h)-f(a))/h​


The Attempt at a Solution


In my attempt I got two different values (the same integer value, except one's a negative).

i) (0-(x-x^3))/(x-1)= (-x+x^3)/(x-1) <-- Wrong substitution



You mixed up what should be f(x) and what should be f(1) in the numerator above.

Then I tried the other equatin:

lim(h->0)- (f(a+h)-f(a))/h

since 1 is the limit, we put 1 in as the a and it equals:

((1+h-(h+3)^3)-(1-1^3))/h <-- Where did h+3 come from?

Again, you have an incorrect substitution in the second form of the derivative. Be more careful in inserting the correct values in the formulas.
 
  • #3
SteamKing said:
You mixed up what should be f(x) and what should be f(1) in the numerator above.

Oh, thanks! That should straighten out the problem.

SteamKing said:
Again, you have an incorrect substitution in the second form of the derivative. Be more careful in inserting the correct values in the formulas.

The "h+3" was a typo. I ended up with the same result. Appreciate it!
 

FAQ: I get two different values when calculating the tangent of a curve.

1. Why am I getting two different values when calculating the tangent of a curve?

This could be due to a few reasons. First, make sure you are using the correct formula for calculating the tangent of a curve. Additionally, if the curve is not a smooth, continuous function, you may get different values at different points. Another factor could be the precision of your calculations, which could result in slight variations in the values.

2. Which value should I use when calculating the tangent of a curve?

Generally, it is recommended to use the average of the two values you get when calculating the tangent of a curve. This can help to reduce any potential errors or variations in the calculations.

3. Is it possible to get more than two values when calculating the tangent of a curve?

No, the tangent of a curve is a unique value at each point on the curve. So, you should only get two values when calculating it – one from the left side and one from the right side of the point.

4. Can the two different values when calculating the tangent of a curve be equal?

Yes, it is possible for the two values to be equal. This could happen if the curve has a point of inflection, where the tangent line is horizontal and has a slope of 0.

5. How can I determine the correct value when calculating the tangent of a curve?

To determine the most accurate value, you can use the limit definition of the derivative. This involves taking the limit as the distance between the two points on the curve approaches 0. Alternatively, you can use graphing software to visually see the tangent line and its slope at a specific point on the curve.

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