Find the X coordinate with slope of -2

[smith007]

Homework Statement

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.

Homework Equations

Fundemental Theorem of Calculus

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 separate 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. :)

 

[gabbagabbahey]

Homework Statement

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.

...

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 separate 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.

Assuming you mean: [tex]y(x)\equiv \int_0^{x^2} \frac{1}{1-\sqrt{t}+t}dt[/tex] and y'(x)=2x or 2x/(1+2x) , then you have the right idea. However, [tex]1-\sqrt{x^2}+x^2=1-|x|+x^2[/tex] 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?

 

[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.

 

[HallsofIvy]

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:

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.

-2 = 2x / 1 - x + x²

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

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)?

 

[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.