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January 12th, 2019, 03:35 PM   #11
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Quote:
Originally Posted by zylo View Post
No problem. Refer to picture in my previous post. Determine tangent from left to right curve and then find the tangent for the center curve using the remaining parameter a.

Back to the problem.

You have a line y=mx+b and two conditions, tangency at left and right curves, to determine m and b. Then find tangent center curve using remaining parameter a.

Equation of center curve which satisfies BC's (-2,-8} and (0,0)
$\displaystyle y=ax^{2}+(2a+4)x =mx+b$
$\displaystyle y'= 2ax+2a+4=m$ and solve for a.

Two eqs in two unknowns: x and a.
At that point the problem was solved. Matching the center curve at the boundaries for continuity is trivial and obvious. But thanks for running out the algebra, which it looks like you may have done from a brief perusal. I assumed once you know what to do, anyone could find the tangent between two curves and I personally wouldn't bother with it unless OP asked for it.

EDIT: Tangent between two curves:

Left curve: $\displaystyle y=x^{2}+10x+8$
At point of tangency: $\displaystyle x_{1},y_{1}$
$\displaystyle y_{1}=x_{1}^{2}+10x_{1}+8$
$\displaystyle y_{1}=mx_{1}+b$
$\displaystyle m=2x_{1}+10$

Right Curve: $\displaystyle y=x+x^{2}$
At point of tangency:$\displaystyle x_{2},y_{2}$
$\displaystyle y_{2}=x_{2}+x_{2}^{2}$
$\displaystyle y_{2}=mx_{2}+b$
$\displaystyle m=1+2x_{2}$

Which are six equations in the six unknowns $\displaystyle x_{1},y_{1},x_{2},y_{2},m,b$.

Last edited by skipjack; January 12th, 2019 at 08:44 PM.
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January 12th, 2019, 09:22 PM   #12
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Quote:
Originally Posted by zylo View Post
I personally wouldn't bother with it unless OP asked for it.
Without doing it, you wouldn't know the values of your m and b, which are needed to "solve for a" as given in your quoted part of your previous post.

Whilst your methods are now correct, you made a slip (that I'll leave you to spot) in setting up the six equations.

Quote:
Originally Posted by JeffM1 View Post
But f(x) consists of 3 segments, each apparently quadratic in form. The tangent of a true quadratic intersects that quadratic only at the point of tangency. Because there is a single line that is tangent to f(x) in 3 distinct places, there must be one point of tangency in each segment.
The original problem didn't state that the tangent touches the graph of f(x) in only 3 places and didn't define "touches". Hence the second segment could be a straight line segment that is technically a tangent to itself at every point in it.
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January 13th, 2019, 06:15 AM   #13
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Quote:
Originally Posted by skipjack View Post
Without doing it, you wouldn't know the values of your m and b, which are needed to "solve for a" as given in your quoted part of your previous post.

Whilst your methods are now correct, you made a slip (that I'll leave you to spot) in setting up the six equations.


The original problem didn't state that the tangent touches the graph of f(x) in only 3 places and didn't define "touches". Hence the second segment could be a straight line segment that is technically a tangent to itself at every point in it.
I actually thought about that, but quickly rejected it. A straight line with limits as indicated would have a slope of

$\dfrac{-\ 8 - 0}{-\ 2 - 0} = 4$, but would necessarily

have to have the equation $y = 4x$ for

$\displaystyle \lim_{x \rightarrow 0} y = 0.$

And the slope of $x^2 + 2x$ at 0 is 0, not 4.

I tried to sneak around this issue with "apparently quadratic" and a "true quadratic," but you caught me fair and square.

Last edited by JeffM1; January 13th, 2019 at 06:19 AM.
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January 13th, 2019, 07:13 AM   #14
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The slope of $x^2 + 2x$ at 0 is 2, not 0. The fact that it isn't 4 doesn't matter.
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January 13th, 2019, 08:52 AM   #15
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The slope of $x^2 + 2x$ at 0 is 2, not 0. The fact that it isn't 4 doesn't matter.
Yes, it is 2.

The fact that it is not 4 does matter.

The tangent to a parabola intersects the parabola exactly once (at the point of tangency obviously) and obviously has the same slope as the derivative of the parabola at the point of tangency. Therefore if a line intersects a parabola at a point and does not have a slope equal to the derivative at that point, it is not a tangent to the parabola.

I conceded to you already that I did not directly address the fact that conceivably the middle segment might be a straight line, but rather indicated indirectly how to dismiss that conceivability.

If the middle segment is a straight line, then the continuity of f(x) requires that the only straight line possible is described by the line y = 4x. If that line is tangent to x^2 + 2x, it must be at a unique point of intersection and the parabola's derivative must equal 4 at that unique point of intersection. To show that the line is not a tangent, you can show either that the derivative is not 4 (which, despite my careless error in my 2nd post, we both agree it is not) or that there is a second point of intersection. The logic is:

$\{A \implies B \text { and } C \} \iff \{ \neg (B \text { and } C) \implies \neg A\} \iff \{ \neg B \text { or } \neg C \implies \neg A\}.$

Perhaps I am making a blunder. If so, it would be a kindness to show me where rather just saying that I am wrong.

EDIT: Is it possible that you are just peeved because I said I did not immediately get your solution for the points of tangency in the outer segments? I did not say your solution was wrong; indeed I said it was a contribution. Did you perhaps read that as sarcasm? It was not. I was saying that the OP might have difficulty seeing how that contribution was derived. My main concerns about this thread were that zylo's posts were likely to confuse the OP and that, given zylo's intervention, the hints given by you and Archie might not be sufficient. My purpose at these sites is to help OP's, not to try to score points against people who actually know much, much more mathematics than I do.
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