Finding Horizontal Tangents

  • Thread starter Hypnos_16
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  • #1
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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.
 

Answers and Replies

  • #2
Dick
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The slope of the tangent is the derivative, right?
 
  • #3
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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
Dick
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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
Mentallic
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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
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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
Mentallic
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So the points where it's horizontal would be
(1 , 0) and (1/3, 16/27)
Yep.

????
Do more examples so you are more confident in yourself when doing these.
 
  • #8
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Thank for all your help man!
 
  • #9
Mentallic
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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
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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
Mentallic
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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.
 

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