Find Equations of tangent lines
I'm having a hard time figuring this out.
I'm asked to find the the equations of the horizontal lines to the curve of
$$y=x^3-3x+1$$
I set the derivative equal to zero and solve for x, to find the constant of each equation (since they are horizontal), I get $-1$ and $1$. When graphing the function, the $-1$ makes sense graphically, but the $1$ doesn't make sense, I realized that even more when asked to find the equations of the lines perpendicular to the above tangent lines at the point of tangency. When looking for the point of tendency it seems I should be getting $y=3$ instead of $y=1$ (at the top of the bell curve)
I'm completely stuck and wonder what I did wrong.
Thanks a lot for your help and sorry about the formatting, not sure why it didn't work.
prompt
graph
calculus curves tangent-line
edited Dec 8 at 17:45
gimusi
1
asked Dec 8 at 17:39
Minimorum
175
add a comment |
I'm having a hard time figuring this out.
I'm asked to find the the equations of the horizontal lines to the curve of
$$y=x^3-3x+1$$
I set the derivative equal to zero and solve for x, to find the constant of each equation (since they are horizontal), I get $-1$ and $1$. When graphing the function, the $-1$ makes sense graphically, but the $1$ doesn't make sense, I realized that even more when asked to find the equations of the lines perpendicular to the above tangent lines at the point of tangency. When looking for the point of tendency it seems I should be getting $y=3$ instead of $y=1$ (at the top of the bell curve)
I'm completely stuck and wonder what I did wrong.
Thanks a lot for your help and sorry about the formatting, not sure why it didn't work.
prompt
graph
calculus curves tangent-line
edited Dec 8 at 17:45
gimusi
1
asked Dec 8 at 17:39
Minimorum
175
add a comment |
I'm having a hard time figuring this out.
I'm asked to find the the equations of the horizontal lines to the curve of
$$y=x^3-3x+1$$
I set the derivative equal to zero and solve for x, to find the constant of each equation (since they are horizontal), I get $-1$ and $1$. When graphing the function, the $-1$ makes sense graphically, but the $1$ doesn't make sense, I realized that even more when asked to find the equations of the lines perpendicular to the above tangent lines at the point of tangency. When looking for the point of tendency it seems I should be getting $y=3$ instead of $y=1$ (at the top of the bell curve)
I'm completely stuck and wonder what I did wrong.
Thanks a lot for your help and sorry about the formatting, not sure why it didn't work.
prompt
graph
calculus curves tangent-line
edited Dec 8 at 17:45
gimusi
1
asked Dec 8 at 17:39
Minimorum
175
I'm having a hard time figuring this out.
I'm asked to find the the equations of the horizontal lines to the curve of
$$y=x^3-3x+1$$
I set the derivative equal to zero and solve for x, to find the constant of each equation (since they are horizontal), I get $-1$ and $1$. When graphing the function, the $-1$ makes sense graphically, but the $1$ doesn't make sense, I realized that even more when asked to find the equations of the lines perpendicular to the above tangent lines at the point of tangency. When looking for the point of tendency it seems I should be getting $y=3$ instead of $y=1$ (at the top of the bell curve)
I'm completely stuck and wonder what I did wrong.
Thanks a lot for your help and sorry about the formatting, not sure why it didn't work.
prompt
graph
calculus curves tangent-line
calculus curves tangent-line
edited Dec 8 at 17:45
gimusi
1
asked Dec 8 at 17:39
Minimorum
175
edited Dec 8 at 17:45
gimusi
1
asked Dec 8 at 17:39
Minimorum
175
edited Dec 8 at 17:45
gimusi
1
edited Dec 8 at 17:45
gimusi
1
edited Dec 8 at 17:45
gimusi
1
1
asked Dec 8 at 17:39
Minimorum
175
asked Dec 8 at 17:39
Minimorum
175
asked Dec 8 at 17:39
Minimorum
175
175
add a comment |
add a comment |
1 Answer
1
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oldest
votes
Your way seems correct, starting form
$$f(x)=x^3-3x+1 implies f'(x)=3x^2-3=0 implies x=pm 1$$
and then
- $f(1)=-1$
- $f(-1)=3$
and the $2$ horizontal lines are then
$$y=-1, quad y=3$$
answered Dec 8 at 17:43
gimusi
1
Right, but I am confused as why the equation of the derivative equal to zero did not give me 3 as a result, shouldn't it have?
– Minimorum
Dec 8 at 17:45
1
@Minimorum No it does not. The roots for $f'(x)=0$ give us the 2 critical points with horizontal tangent which are point of maximum and point minimum in that case . We need to plug the values into $f(x)$ to find the corresponding $y$ values.
– gimusi
Dec 8 at 17:49
Ooooh got it. Makes sense. For some reason I was getting stuck on the initial roots. Thanks so much for the help
– Minimorum
Dec 8 at 17:51
@Minimorum The point on f(x) with horizontal tangent line are then $(1,-1)$ and $(-1, 3)$.
– gimusi
Dec 8 at 17:51
@Minimorum You are welcome! You doubt was a simple issu but you have presented that in a proper way with all the details. Well done! Bye
– gimusi
Dec 8 at 17:52
add a comment |
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Your way seems correct, starting form
$$f(x)=x^3-3x+1 implies f'(x)=3x^2-3=0 implies x=pm 1$$
and then
- $f(1)=-1$
- $f(-1)=3$
and the $2$ horizontal lines are then
$$y=-1, quad y=3$$
answered Dec 8 at 17:43
gimusi
1
Right, but I am confused as why the equation of the derivative equal to zero did not give me 3 as a result, shouldn't it have?
– Minimorum
Dec 8 at 17:45
1
@Minimorum No it does not. The roots for $f'(x)=0$ give us the 2 critical points with horizontal tangent which are point of maximum and point minimum in that case . We need to plug the values into $f(x)$ to find the corresponding $y$ values.
– gimusi
Dec 8 at 17:49
Ooooh got it. Makes sense. For some reason I was getting stuck on the initial roots. Thanks so much for the help
– Minimorum
Dec 8 at 17:51
@Minimorum The point on f(x) with horizontal tangent line are then $(1,-1)$ and $(-1, 3)$.
– gimusi
Dec 8 at 17:51
@Minimorum You are welcome! You doubt was a simple issu but you have presented that in a proper way with all the details. Well done! Bye
– gimusi
Dec 8 at 17:52
add a comment |
Your way seems correct, starting form
$$f(x)=x^3-3x+1 implies f'(x)=3x^2-3=0 implies x=pm 1$$
and then
- $f(1)=-1$
- $f(-1)=3$
and the $2$ horizontal lines are then
$$y=-1, quad y=3$$
answered Dec 8 at 17:43
gimusi
1
Right, but I am confused as why the equation of the derivative equal to zero did not give me 3 as a result, shouldn't it have?
– Minimorum
Dec 8 at 17:45
1
@Minimorum No it does not. The roots for $f'(x)=0$ give us the 2 critical points with horizontal tangent which are point of maximum and point minimum in that case . We need to plug the values into $f(x)$ to find the corresponding $y$ values.
– gimusi
Dec 8 at 17:49
Ooooh got it. Makes sense. For some reason I was getting stuck on the initial roots. Thanks so much for the help
– Minimorum
Dec 8 at 17:51
@Minimorum The point on f(x) with horizontal tangent line are then $(1,-1)$ and $(-1, 3)$.
– gimusi
Dec 8 at 17:51
@Minimorum You are welcome! You doubt was a simple issu but you have presented that in a proper way with all the details. Well done! Bye
– gimusi
Dec 8 at 17:52
add a comment |
Your way seems correct, starting form
$$f(x)=x^3-3x+1 implies f'(x)=3x^2-3=0 implies x=pm 1$$
and then
- $f(1)=-1$
- $f(-1)=3$
and the $2$ horizontal lines are then
$$y=-1, quad y=3$$
answered Dec 8 at 17:43
gimusi
1
Your way seems correct, starting form
$$f(x)=x^3-3x+1 implies f'(x)=3x^2-3=0 implies x=pm 1$$
and then
- $f(1)=-1$
- $f(-1)=3$
and the $2$ horizontal lines are then
$$y=-1, quad y=3$$
answered Dec 8 at 17:43
gimusi
1
edited Dec 8 at 17:52
answered Dec 8 at 17:43
gimusi
1
answered Dec 8 at 17:43
gimusi
1
answered Dec 8 at 17:43
gimusi
1
1
Right, but I am confused as why the equation of the derivative equal to zero did not give me 3 as a result, shouldn't it have?
– Minimorum
Dec 8 at 17:45
1
@Minimorum No it does not. The roots for $f'(x)=0$ give us the 2 critical points with horizontal tangent which are point of maximum and point minimum in that case . We need to plug the values into $f(x)$ to find the corresponding $y$ values.
– gimusi
Dec 8 at 17:49
Ooooh got it. Makes sense. For some reason I was getting stuck on the initial roots. Thanks so much for the help
– Minimorum
Dec 8 at 17:51
@Minimorum The point on f(x) with horizontal tangent line are then $(1,-1)$ and $(-1, 3)$.
– gimusi
Dec 8 at 17:51
@Minimorum You are welcome! You doubt was a simple issu but you have presented that in a proper way with all the details. Well done! Bye
– gimusi
Dec 8 at 17:52
add a comment |
Right, but I am confused as why the equation of the derivative equal to zero did not give me 3 as a result, shouldn't it have?
– Minimorum
Dec 8 at 17:45
1
@Minimorum No it does not. The roots for $f'(x)=0$ give us the 2 critical points with horizontal tangent which are point of maximum and point minimum in that case . We need to plug the values into $f(x)$ to find the corresponding $y$ values.
– gimusi
Dec 8 at 17:49
Ooooh got it. Makes sense. For some reason I was getting stuck on the initial roots. Thanks so much for the help
– Minimorum
Dec 8 at 17:51
@Minimorum The point on f(x) with horizontal tangent line are then $(1,-1)$ and $(-1, 3)$.
– gimusi
Dec 8 at 17:51
@Minimorum You are welcome! You doubt was a simple issu but you have presented that in a proper way with all the details. Well done! Bye
– gimusi
Dec 8 at 17:52
Right, but I am confused as why the equation of the derivative equal to zero did not give me 3 as a result, shouldn't it have?
– Minimorum
Dec 8 at 17:45
Right, but I am confused as why the equation of the derivative equal to zero did not give me 3 as a result, shouldn't it have?
– Minimorum
Dec 8 at 17:45
1
1
@Minimorum No it does not. The roots for $f'(x)=0$ give us the 2 critical points with horizontal tangent which are point of maximum and point minimum in that case . We need to plug the values into $f(x)$ to find the corresponding $y$ values.
– gimusi
Dec 8 at 17:49
@Minimorum No it does not. The roots for $f'(x)=0$ give us the 2 critical points with horizontal tangent which are point of maximum and point minimum in that case . We need to plug the values into $f(x)$ to find the corresponding $y$ values.
– gimusi
Dec 8 at 17:49
Ooooh got it. Makes sense. For some reason I was getting stuck on the initial roots. Thanks so much for the help
– Minimorum
Dec 8 at 17:51
Ooooh got it. Makes sense. For some reason I was getting stuck on the initial roots. Thanks so much for the help
– Minimorum
Dec 8 at 17:51
@Minimorum The point on f(x) with horizontal tangent line are then $(1,-1)$ and $(-1, 3)$.
– gimusi
Dec 8 at 17:51
@Minimorum The point on f(x) with horizontal tangent line are then $(1,-1)$ and $(-1, 3)$.
– gimusi
Dec 8 at 17:51
@Minimorum You are welcome! You doubt was a simple issu but you have presented that in a proper way with all the details. Well done! Bye
– gimusi
Dec 8 at 17:52
@Minimorum You are welcome! You doubt was a simple issu but you have presented that in a proper way with all the details. Well done! Bye
– gimusi
Dec 8 at 17:52
add a comment |
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