# Homework Help: Tangent problem doubt.

1. Aug 11, 2011

### naaa00

Hello people!

Well, I am doing some excercises for fun. Picked some Precalculus stuff, and found this excercise: "Construct a function that has the same slope at x = 1 and x = 2. Then find two points where y = x^4 - 2x^2 has the same tangent line (draw the graph)." I have found a solution, but I think the solution is totally wrong. Any advice?

What I did:

For the first part made up y = x^3 + 1

Then, I plotted two points: (1,2) (2,9) where (a,f(a)) and (b,f(b)).

Calculated the slope and got 7.

two-point form: y - 2=7(x - 1)
y = 7x - 9 slope-form. Secant line.
y= 7 Tangent line.

For the second par did something similar:

I plotted points: (2,24) (3,81) for y = x^4 - 2x^2

The slope is 57.

Thw two-point form: y -24 =57(x - 2)
y = 57x - 90 slope-form. Secant line.
y = 57 Tangent line.

It must be wrong, but I can not say why.

Also tried the following:

Since two points have the same tangent line, or derivative, f'(a) = f'(b)

And y = x^4 - 2x^2
when x = a
y = a^4 -2a^2 and y = b^2 - 2b^2

Then: f'(a) = f(a) - f(b)/ a - b
f'(b) = f(b) - f(a)/ b - a

End up needing to solve for two equations, but everything gets to messy, and surely must be wrong.

I think I am confused.

Any correction?

Last edited: Aug 11, 2011
2. Aug 11, 2011

### Bohrok

What you did was find two point from the function y = x3 + 1 and find that the secant line between them had a slope of 7. Where was the tangent line y = 7 coming from?
A function that has the same slope at x = 1 and x = 2 means that its derivative at x = 1 and x = 2 are the same.

For the second problem, you want to find two points on y = x4 - 2x2 where the derivative at the two points are the same.

3. Aug 11, 2011

### HallsofIvy

If $y= x^3+ 1$, then $y'= 3x^2$. At x= 1, the slope is y'= 3 while at x= 2, the slope is y'= 12. Those are NOT the "same slope"

4. Aug 11, 2011

### naaa00

Thank you for your answer Bohrok and HallsofIvy. Well, How did I got y = 7? Actually it is a little bit stupid. I though the following:

A) y - a = f(c) - f(a)/c-a (x-a) Secant line

B) y - a = f'(a)(x-a) for the limit of the secant line.

Basicaly I differentiate the 7x to 7, since nx^n-1... Equation "A" I saw it as the average rate of change and "B" as the limit, as c approaches a.

Indeed. Then, my answer is very local? Or is simply wrong?

5. Aug 12, 2011

### naaa00

Tonigh while I was drinking an horrible apple juice in some horrible place I was able to see my wrong understanding of the concepts! I will try to do it again tomorrow, from the begining. I'll let you know guys how it goes. Thanks in advanced!

6. Aug 12, 2011

### Staff: Mentor

The instruction to draw the graph is there for a reason. A good reason--so that you can see what you are doing. http://fooplot.com/index.php?q0=x^4-2*x^2

7. Aug 12, 2011

### HallsofIvy

Yes, horrible apple juice will do that- depending upon just how much it has fermented.

8. Aug 15, 2011

### naaa00

Ha, ha, ha. INDEED. Well, I am back.

I answered the problem. But it is TOO simple to be true:

Well, I just saw the graph, and realized that there were two points where the slope is zero (no change). Those points are (1,-1) and (-1,-1). Those points have the same slope... I found those points making a table of values. (I don't like to use graphing calculators, but did payed attention to NascentOxygen's suggestion)

That could be considered a solution, but I don't like it. It strikes me as an easy way to solve the problem. Is it the only way to solve it? Perhaps in this case, but let's suppose some function where there is an equal rate of change at to distinct points and their derivatives is not equal to zero. Any ideas?

For part a a similar question arises. I could easily construct a function such that y(x) is equal to some constant. But, again, the same problem arises, for we will have the same slope for all x. What if, if I would like a function that is the same at x = 1 and x = 2, but different at any other point?

I am not satisfied with my answers. Any ideas?

Last edited: Aug 15, 2011
9. Aug 15, 2011

### PeterO

I think you know a quadratic [highest power of x is 2] is a parabola. Just by looking at the shape you can see that no two points on the parabola have the same gradient.

A cubic [highest power of x =3] has a graph that looks a little like an S on it side. If you sketch the cubic, you will find pairs of places where parallel tangents can be drawn to the cubic [that was the first question]

A Quartic function [highest power of x = 4] can look like a rounded off W. Depending on the complexity of the quartic, the two "bottoms" of the quartic may be at different heights, like the Southern coast of Australia where the low point in Western Australia is much closer to the equator than the low point of Eastern Australia [Victoria]

It is possible to draw a single tangent that touches both low points [question 2].

All you have to do now is locate those points etc.
Note, at those points, the gradient will be the same

10. Aug 15, 2011

### naaa00

Thank you for your answer, PeterO! I knew that there was something incomplete in my understanding of the essence of the problem. I will now try to resolve similar problems, but more complicated than this one. Thank you!

11. Aug 15, 2011

### Staff: Mentor

If you examine the plot I linked to earlier, viz., http://fooplot.com/index.php?q0=x^4-2*x^2 you can see wide regions of similar slopes. For example, in the region -1<x<-0.5 you find every slope between 0 and +45o; and you can see an identical range of slopes in the region -0.5<x<0 and the same again in the region 1<x<1.1

12. Aug 15, 2011

### naaa00

Aha, I see...

(Now I am zomming like crazy all the points in the graph) But it seems to me, then, that there can be MANY (much more than I though) points with equal slope... This is anarchy.

13. Aug 15, 2011

### PeterO

There is an infinite number!!!!

14. Aug 15, 2011

### naaa00

Now it turns out that the tangent line is worst than a prostitute: one point OR infinitely many points? Infinitely many points... Is not the definition a little bit deceiving?
"Suppose the motion of an asteroid whose track follows f(x) = x^4-2*x^2+x+1. At point P, the asteroid was hit by a missile, and as a consecuence, one part of the asteroid is detached of it. The detached part crashes the asteroid at a second point. Write the equation of the track of the detached part."

So this, according to PeterO, is solvable, if we choose that the tangent line passes through both points.(?)

15. Aug 15, 2011

### PeterO

What makes you think the bit broken off "leaves tangentially", and for that matter, makes a tangential glancing blow with some other part of the asteroid, or its path????

16. Aug 15, 2011

### naaa00

Well, I am asuming that that happened for some unknown and mysterious reason, but that is a peculiar assumption that I took for the purposes of the topic in question. It may have no sense at all as many 2D physics classical mechanics problems. The context of the problem is not "real life" but, say, "Naaa00's space", where its laws of motion make it happen.

I though that maybe it could work as an excercise, (as I said, for the purposes of the topic in question, or?) Perhaps the asteroid is not a very good example for expressing my question, so instead of an asteroid, let a billard ball be our instance:

"Suppose the track that follows the motion of an billard ball (ball number 3, or "b3") after being hit by another ball is f(x) = x^4-2*x^2+x+1, At point P, "b3" hit another ball (ball 8). but "b8" hits again "b3" at a second point. Write the equation of the track of ball 8."

Assuming that the track of b8 "leaves tangentially", I am supposed to be able to come up with the correct equation, or? I think so. Or am I fantasizing?

17. Aug 15, 2011

### PeterO

Try another example - billiard balls travel in straight lines - unless you have a complex un-flat table.

18. Aug 15, 2011

### PeterO

Breaking down you original post to separate the first task.

What you did was make up a function that did NOT meet the requirements.

The simplest function with two points having the same gradient is a cubic.

You need the function, and the gradient function to work with - that is the cubic function and its quadratic derivative. [like x^3 and 3x^2 - but they are not suitable]

You could have chosen any two points with any gradient and solved the problem, but it may be even easier if you choose those two points to have zero gradient.

I would chose the points to be something like (1 , 3) and (2 , 0) and say that the gradient at each point is zero. [note: the only restriction here is that the y ordinate is not the same for each or it won't work - make them equal and you will encounter a problem later on.

Generally

y = ax^3 + bx^2 + cx + d ... (1)

y' = 3ax^2 + 2bx + c ... (2)

The (x,y) pairs (1,3) and (2,0) are substituted into equation (1) to generate 2 equations in a,b,c&d
The (x,y') pairs (1,0) and (2,0) are substituted into equation (2) to generate 2 more equations in a,b&c.
Four equations, 4 unknowns - they can be solved.

Note: equation (1) is the function; equation (2) is the gradient function.

19. Aug 15, 2011

### naaa00

Thank you for your explanation, PeterO! Much appreciated. So basicaly I can do whatever I want. That's quite interesting.

So, If I want to create a function, say, for a problem that I want to do from scratch for fun, to create it, it would be something similar to reverse engineering. At least that's how I feel it.

About the billard balls example... Epic fail from my part...

20. Aug 16, 2011

### PeterO

Generally it is not so much a free for all, as they will specify not only that the gradient is the same, but also what gradient is supposed to be. Perhaps even a y-intercept will be thrown in.

As for the quartic x^4 - x^2. That can be factorised to x^2(x^2 - 1) = x^2(x-1)(x+1)

Now y = x^2(x-1)(x+1) could perhaps be re-written as y = (x-0)(x-0)(x-1)(x+1) to emphasise that there is a repeated root at x = 0, and other roots at x = 1 and x = -1.

If you were to draw a graph of this it would be a "rounded off W".

It would cut the x -axis at -1 and +1 and touch the x axis at the origin [it would cut the y-axis at the origin.

You would be able to draw a horizontal line, slightly below the x-axis which would touch the twin low p;oints of the graph.

That is the line they were after in part 2.

The equatio of that line would be y = k where k is a number, independent of x