How do I find points on a curve where the tangent is horizontal?

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In summary, the Homework statement is trying to find the points on the curve y=x3−x2−x+1 where the tangent is horizontal. The Attempt at a Solution does not have a solution, and states that for it to be horizontal, the slope of the tangent has to equal 0.
  • #1
Hypnos_16
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Homework Statement



Find the points on the curve y=x3−x2−x+1 where the tangent is horizontal.
I know that for it to be horizontal it needs a slope of 0

Homework Equations



y = mx + b

The Attempt at a Solution



I don't have one, i can't think where to even start with this.
 
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  • #2
The slope of the tangent is the derivative, right?
 
  • #3
yes, and i know that for it to be horizontal, that the slope has to equal 0. But that's the part that stumps me. I assume I'm not supposed to just fill in values till i get an answer of 0. It just doesn't make sense to me.
 
  • #4
Hypnos_16 said:
yes, and i know that for it to be horizontal, that the slope has to equal 0. But that's the part that stumps me. I assume I'm not supposed to just fill in values till i get an answer of 0. It just doesn't make sense to me.

I'm not sure why it's not making sense. Take the derivative and set it equal to zero. That gives you a quadratic equation in x. You've solved quadratics before I'll bet.
 
  • #5
If you have a curve y=f(x), then its derivative dy/dx will tell you the gradient of the tangent that touches the curve at the point x. If you set dy/dx=1 then when you solve for x, you will find all x values where the gradient of the tangent on the curve f(x) is equal to 1.
Since you want horizontal tangents, you want to set the derivative equal to 0.

So firstly, what's the derivative?
 
  • #6
The Derivative of y = x3 - x2 - x + 1
is F'(x) = 3x2 - 2x - 1
So then, i set F'(x) to 0
getting 3x2 - 2x - 1 = 0
3x2 - 2x - 1 = 0

AFTER QUADRATIC.
(x = 1)(x = -1/3)

y = (1)3 - (1)2 - (1) + 1
y = 1 - 1 - 1 + 1
y = 0

y = (1/3)3 - (1/3)2 - (1/3) + 1
y = (1 / 27) - (1 / 9) - (1 / 3) + 1
y = (1 / 27) - (3 / 27) - (9 / 27) + (27 / 27)
y = (16 / 27)

So the points where it's horizontal would be
(1 , 0) and (1/3, 16/27)
?
 
  • #7
Hypnos_16 said:
So the points where it's horizontal would be
(1 , 0) and (1/3, 16/27)
Yep.

Hypnos_16 said:
?
Do more examples so you are more confident in yourself when doing these.
 
  • #8
Thank for all your help man!
 
  • #9
Just curious, but I'm finding it kind of odd that you're posting two questions that (in my opinion) have a big difference in their difficulty level. This problem is much easier than your other one.
 
  • #10
Agreed. It's some assignment I'm practicing , it's difficulty level fluctuates all over the place. I get all the ones i can get done, done. They i reread my notes and check the text. Then if there is still no prevail, i try here. I don't like using it, cause i want to get it on my own. But these have driven me crazy.
 
  • #11
Ahh a past exam, that makes sense then :smile: Good luck with it!
And by the way, spend a good hour understanding the process of answering questions like these. You've probably taken enough derivatives by now, the process is what matters the most at this point.
 

1. What is the definition of a horizontal tangent?

A horizontal tangent is a line that is tangent to a curve at a specific point and is parallel to the x-axis at that point.

2. How do you find the coordinates of a point where a curve has a horizontal tangent?

To find the coordinates of a point where a curve has a horizontal tangent, you must first take the derivative of the curve and set it equal to 0. Then, solve for the x-value(s) that make the derivative equal to 0. These x-values will be the coordinates of the point(s) where the curve has a horizontal tangent.

3. Can a curve have more than one point where it has a horizontal tangent?

Yes, a curve can have multiple points where it has a horizontal tangent. This occurs when the derivative of the curve is equal to 0 at more than one x-value.

4. How do you know if a curve has a horizontal tangent at a specific point?

A curve has a horizontal tangent at a specific point if the derivative of the curve is equal to 0 at that point. This means that the slope of the curve at that point is 0, resulting in a horizontal tangent line.

5. What is the significance of finding horizontal tangents in real-life applications?

Finding horizontal tangents can help determine the maximum or minimum points of a curve, which can be useful in optimization problems. It can also be used in physics to find points where an object is at rest or has constant velocity. Additionally, horizontal tangents can be used in economics to find the break-even point for a business.

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