# Math Help - tangent proofs for y=x^3

1. ## tangent proofs for y=x^3

Hey everyone,

I recently decided to reopen to old uni maths book and work it through from cover to cover, completing every single problem, proof etc. It is taking a while, but enjoying it. I am sure I will get stuck again after this at some point, so thanks in advance everyone! Actually supposed to be working right now, but spending an hour every morning on this stuff

Anyway stuck on the following problem (R. A. Adams, Calculus a Complete Course, 5th Edition, Ch 2, ch review prob 15 if you want to know ):

Let C be the graph of y = x^3
a) show that if a /=0 then the tangent to C at x=a also intersects C at a second point x=b
b) show that the slope of C at x=b is four times its slope at x=a
c) Can any line be tangent to C at more than one point
d) Can any line be tangent to te graph of y = Ax^3 + Bx^2 + Cx + D

I have not gotten very far, but this is what I have at present:

y' = 3x^2
let x = a, therefore the tangent is: y=3a^2(x-a)+a^3 = 3a^2x - 2a^3
For a) I assume that C(b) means that y = b^3 for x=b

Then I get stuck...

2. ## Re: tangent proofs for y=x^3

Let C be the graph of y = x^3
a) show that if a /=0 then the tangent to C at x=a also intersects C at a second point x=b
b) show that the slope of C at x=b is four times its slope at x=a
c) Can any line be tangent to C at more than one point
d) Can any line be tangent to te graph of y = Ax^3 + Bx^2 + Cx + D
(a) at $(a,a^3)$ , tangent slope is $3a^2$

if this tangent line intersects the curve again at $(b,b^3)$ , $a \ne b \ne 0$ , then

$3a^2 = \frac{b^3-a^3}{b-a} = b^2 + ab + a^2 \implies b^2 + ab - 2a^2 = 0$

$b^2 - a^2 + ab - a^2 = 0$

$(b+a)(b-a) + a(b-a) = 0$

since $a \ne b$ ...

$(b-a)(b + 2a) = 0 \implies b = -2a$

(b) $y'(a) = 3a^2$

$y'(b) = 3b^2 = 3(-2a)^2 = 12a^2 = 4 \cdot 3a^2 = 4 \cdot y'(a)$

(d) "any line" ??? I would say no ... how about the vertical line $x = k$? or any horizontal line where $3Ax^2 + 2Bx + C \ne 0$

...

4. ## Re: tangent proofs for y=x^3

Thanks for the help, my other reply vanished somehow

For c) how does the answer to b) answer this? Would you mind giving me a little explaination?

5. ## Re: tangent proofs for y=x^3

I think I have some kind of proof for d), thoughts?

this curve will only have two solutions for horizontal lines if the equation y = 3Ax^2 + 2xB + C = 0, has two real solutions.
depending on A, B, C
the discriminant = Delta = sqrt(4B^2 - 3AC)
case 1, Delta < 0: no real solutions, therefore the critical point is an inflection, and the tangent at this point will only be tangent at the critical point and will not be horizontal, since that is impossible. i.e. y = x^2 + x + 1
case 2, Delta > 0: two real solutions, therefore two critical points, and the tangent at that point is a horizontal line, i.e. y = x^2 + x - 1
case 3, Delta = 0: one real solution, i.e. the case y = x^3

therefore tangents where y = 3Ax^2 + 2xB + C <= 0 have no real solutions/one solution, will only be tangent at one point.

actually at points of inflection is a tangent even possible? Thinking of limits, the left and right limits should be different?
If this is the case then where delta < 0 has be no point where it is a tangent

therefore there are tangent lines that have no solution (delta<0) or one solution (delta=0)

summary: no, some lines are not tangent to the graph y = Ax^3 + Bx^2 + Cx + D

6. ## Re: tangent proofs for y=x^3

Originally Posted by tamanous
For c) how does the answer to b) answer this? Would you mind giving me a little explaination?
(b) established that the line tangent to any point on the curve intersects the curve at another point on the curve ... if a line is tangent to a curve twice, then it has to intersect the curve at both points and have the same slope, right?

7. ## Re: tangent proofs for y=x^3

Originally Posted by skeeter
(b) established that the line tangent to any point on the curve intersects the curve at another point on the curve ... if a line is tangent to a curve twice, then it has to intersect the curve at both points and have the same slope, right?
Ah, of course, thanks