You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

** Let f be the function defined by f(x) = x^3 - x^2 -4x + 4. The point (a,b) is on the graph of f, and the line tangent to the graph at (a,b) passes through the point (0, -8), which is not on the graph of f. Find a and b.**

Ok, the line tangent to the graph of f is the derivaitve of f.

f(x) = x^3 - x^2 -4x + 4

f'(x) = 3x^2 -2x - 4

f'(x) is merely the slope of the line. The equation of the line tangent to f is:

y = (3x^2 - 2x - 4)x + b

NOTE! b is the y intercept, not the variable b in the problem. We were told that the line tangent to f at the point (a,b) passes through (0,-8). We have not found precisly what the slope is, but no matter what it is, it will have a value of zero at this point. Therefore:

-8 = (3(0^3) - 2(0) - 4)0 + b

thus b equals -8

So we have:

y = (3x^2 - 2x -4)x - 8

what this eqation represents is somewhat abstract. The graph of this equation and the graph of f will intersect at (a,b).

Therefore:

f(x) = (3x^2 - 2x -4)x - 8

x^3 - x^2 -4x + 4 = 3x^3 - 2x^2 -4x - 8

-2x^3 + x^2 + 12 = 0

2x^3 - x^2 - 12 = 0

as far as I can tell this can be simplified no further.

Sure you can solve this using the rational roots theorem and or synthetic division. Or with a graphing calcuator. And the answer it yields is correct. x = 2, so a =2 and b = 0. But this was a somewhat inelligent method of solving the problem I think. I think there must be a simple way to solve it, one that does not end up in a third degree polynomial that cannot be reduced. I've never seen a problem in my mathbook that required the solving of a third degree polynomial and didn't say "use a graphing calculator as an aid for solving the problem".

So, any idea's?

A logarithm is just a misspelled algorithm.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Sorry, I can't think of anything. I thought I had something, but it turned out it was wrong. Then I thought I had another something, but that was wrong too.

I found out something quite weird though. I tried to find out if I could factorise it by doing long division by (x+a), where a is a constant. Then, I planned to look at the remainder and work out for which value of a that would be 0.

Guess what remainder I got. 2a³+a²-12.

Why did the vector cross the road?

It wanted to be normal.

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I guess its just a genuine badass problem. I'll give it another shot.

(runs in the cage to take on the problem, and the ref shuts the door)

POW!

"ooooh... THAT ONE hurt him!" (mikau is carried out on a stretcher)

"NEXT!"

*Last edited by mikau (2005-12-19 15:05:15)*

A logarithm is just a misspelled algorithm.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

2x^3 - x^2 - 12 = 0

This can be factored to:

(x-2)(2x^2+3x+6) = 0

So x=2 is the only real solution (show by using the quadratic formula).

x=2, y(2) = 8 - 4 - 8 + 4 = 0

So the point is (2, 0).

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Yep yep. But like I said, it is not imediatly apparent that it can be factored into that form. If it was, then we could solve the problem right away. It leaves no remainder and thus is a zero of the expression or function when the factor equals 0.

*Last edited by mikau (2005-12-19 18:25:18)*

A logarithm is just a misspelled algorithm.

Offline

**Maths Brainiac****Member**- Registered: 2005-12-19
- Posts: 1

What is the question coz i might able 2 do it?????????????

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,663

Ummm ... the question is the bold bit at the top.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Well, no one said math should be immediately apparent

If you don't want to try to factor, you could always try the cubic formula:

http://mathworld.wolfram.com/CubicFormula.html

So you don't have to have a calculator, but it is much simpler if you do. Isn't that why those were invented anyways?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Is there a method to finding the (x-2) factor of the cubic?

Or did Ricky divide by it because we had already known the root?

I'll have to learn what rational roots theorem and synthetic division is.

*Last edited by John E. Franklin (2005-12-20 11:54:29)*

**igloo** **myrtilles** **fourmis**

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Synthetic division is teh UBER roxxor! And very easy. I just got in from the cold and I can barely type. Later on when I'm comfy I'll write an explanation.

A logarithm is just a misspelled algorithm.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

"Or did Ricky divide by it because we had already known the root?"

I played around with the numbers, trying different factors till I found (x-2) was one. Took me about 10 minutes to do.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

So it was basically trial and error, then. If it wasn't factorisable, you would have been there for quite a while.

Why did the vector cross the road?

It wanted to be normal.

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I was already there for quite a while. Must have spent and hour on the problem. lol.

A logarithm is just a misspelled algorithm.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

mathsyperson wrote:

So it was basically trial and error, then. If it wasn't factorisable, you would have been there for quite a while.

Sort of. Since the coefficients are all low numbers, you know the factors have to be low as well. So you can just try all the low integer factors -5 to 5 and see if they work. If those don't, you can be reasonably sure the factors aren't integers.

Offline

Pages: **1**