Find the X coordinate with slope of -2

1. Dec 10, 2008

smith007

1. The problem statement, all variables and given/known data

Function defined as:

y = The integral from 0 to x2 of 1 / (1 - Sqrt(t) + t)

There exists a point where the slope of the tangeant is = -2. Find the x coordinate at this point.

2. Relevant equations

Fundemental Theorem of Calculus

3. The attempt at a solution

If I use FTC and replace t with x2. Then I get two equations because sqrt(x2) is +/- x. I use those to create two seperate equations.

y = 2x and y = 2x/1+2x

These should be 2 equations for the tangeant line but I am unsure what to do next.

This was on my exam yesterday and I tried some things, none of which seemed right. I know I lost mark just would like to know how many. :)

2. Dec 10, 2008

gabbagabbahey

Assuming you mean: $$y(x)\equiv \int_0^{x^2} \frac{1}{1-\sqrt{t}+t}dt$$ and y'(x)=2x or 2x/(1+2x) , then you have the right idea. However, $$1-\sqrt{x^2}+x^2=1-|x|+x^2$$ not 1 or 1+2x!.....And y'(x) is the equation for the slope of the tangent line, not the tangent line itself!

So, where does y'(x)=-2?

3. Dec 10, 2008

smith007

I forgot one of the squares on the bottom, bad mistake.

So my equation becomes:
y' = 2x / 1 - |x| + x2

-2 = 2x / 1 - |x| + x2 but we know x is positive because of the range of the original question so it becomes:

-2 = 2x / 1 - x + x2

Solve for x = 1? I still don't think that is quite right.

4. Dec 10, 2008

HallsofIvy

Staff Emeritus
Where did this come from? You didn't say anything about a range or domain (which is what I think you really mean) before. The only reference to the variable was a "x2 in which x can be positive of negative.

That isn't at all right because x= 1 does NOT satify that equation!

But if you take x to be negative then you must have -2= 2x/(1+ x+ x2) so that -1- x- x2= x or x2+ 2x+ 1= 0. What negative solution does that have? Does it satisfy -2= 2x/(1- |x|+ x2)?

5. Dec 10, 2008

gabbagabbahey

As Halls said, unless you are explicitly told that x must be positive, you must examine two cases: (1)x is positive and (2)x is negative.