Find equation of line f(x) parallel to given line

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July 30th 2009, 03:48 PM
drkidd22

Find equation of line f(x) parallel to given line

This is another one of those problems that I kinda get the idea of how to do it but then just can't think of anything else. So, to the problem.

Function
$ f(x) = x^3 $
$f'(x) = 3x^2$

Line
$ 3x-y+1 = 0 $
$y = 3x+1$

From here we know that $m=3$ since $y=mx+b$

Then I made $f'(x)=3$
$3x^2=3$
$x^2=1$

To get the value of x.
$x=1$

Now I go back to the given line and plug in:
$y=3(1)+1$
$y=4$

My point is $(1,4)$

I did most of it as I was writing this, I think my light bulb turned on.

$y-y1=m(x-x1)$
$y-4=3(x-1)$
$y=3x+1$

Did I do it right or is there another way of doing this?
July 30th 2009, 03:58 PM
skeeter

Quote:

Originally Posted by drkidd22

This is another one of those problems that I kinda get the idea of how to do it but then just can't think of anything else. So, to the problem.

Function
$ f(x) = x^3 $
$f'(x) = 3x^2$

Line
$ 3x-y+1 $ is this 3x-y+1 = 0 ?
$y = 3x+1$

From here we know that $m=3$ since $y=mx+b$

Then I made $f'(x)=3$
$3x^2=3$
$x^2=1$

To get the value of x.
$x=1$

x = -1 also

Now I go back to the given line and plug in:
$y=3(1)+1$ 3(1) + 1 = 4, not 5
$y=5$

.
July 30th 2009, 04:03 PM
Haytham

Quote:

Originally Posted by drkidd22

This is another one of those problems that I kinda get the idea of how to do it but then just can't think of anything else. So, to the problem.

Function
$ f(x) = x^3 $
$f'(x) = 3x^2$

Line
$ 3x-y+1 $
$y = 3x+1$

From here we know that $m=3$ since $y=mx+b$

Then I made $f'(x)=3$
$3x^2=3$
$x^2=1$

To get the value of x.
$x=1$

Now I go back to the given line and plug in:
$y=3(1)+1$
y=5

My point is $(1,5)$

I did most of it as I was writing this, I think my light bulb turned on.

$y-y1=m(x-x1)$
$y-5=3(x-1)$
$y=3x+2$

Did I do it right or is there another way of doing this?

red part is wrong
if I understood the question you should enter the $x=1$ in the $ f(x) = x^3 $ function, instead of the line equation. the line equation will give you the line equation, if you correct the red part
July 30th 2009, 04:04 PM
drkidd22

http://www.mathhelpforum.com/math-he...37475451-1.gif is this 3x-y+1 = 0 ? yes

Then I made http://www.mathhelpforum.com/math-he...37b7b5b7-1.gif
http://www.mathhelpforum.com/math-he...a2174a5a-1.gif
http://www.mathhelpforum.com/math-he...05e75f07-1.gif

To get the value of x.
http://www.mathhelpforum.com/math-he...235bd26d-1.gif

x = -1 also why is this?

http://www.mathhelpforum.com/math-he...1bdbdd50-1.gif 3(1) + 1 = 4, not 5 My bad, but then if you say x = -1 it would be 3(-1)+1= -2?
http://www.mathhelpforum.com/math-he...a9ae7958-1.gif
July 30th 2009, 04:08 PM
drkidd22

http://www.mathhelpforum.com/math-he...8f41dd91-1.gif this is the derrivative of $f(x)=x^3$ which you need to find the slope, since the slope is 3 that why I made http://www.mathhelpforum.com/math-he...8f41dd91-1.gif=3 to get x = 1
July 30th 2009, 04:27 PM
skeeter

Quote:

Originally Posted by drkidd22

http://www.mathhelpforum.com/math-he...37475451-1.gif is this 3x-y+1 = 0 ? yes

Then I made http://www.mathhelpforum.com/math-he...37b7b5b7-1.gif
http://www.mathhelpforum.com/math-he...a2174a5a-1.gif
http://www.mathhelpforum.com/math-he...05e75f07-1.gif

To get the value of x.
http://www.mathhelpforum.com/math-he...235bd26d-1.gif

x = -1 also why is this?

if $x^2 = 1$ , then $x^2 - 1 = 0$ ... $(x+1)(x-1) = 0$

this says that there will be two distinct lines tangent to the curve y = x^3 that are parallel to $y = 3x+1$
July 30th 2009, 05:13 PM
drkidd22

I'm not complettly clear on this problem yet. So what is the equation of the line parallel to the given line? and how is it done?
July 30th 2009, 05:26 PM
skeeter

Quote:

Originally Posted by drkidd22

I'm not complettly clear on this problem yet. So what is the equation of the line parallel to the given line? and how is it done?

you've established that the slope of a line parallel to y = 3x+1 is 3.

$f'(x) = 3$

$3x^2 = 3$

$x = \pm 1$

now ... get the two points on the original curve where f'(x) = 3

$y = x^3$

$y = (1)^3$ ... (1,1)

y - 1 = 3(x - 1)

y = 3x - 2 is the first line

$y = (-1)^3$ ... (-1,-1)

y + 1 = 3(x + 1)

y = 3x + 2 is the second line

now look at the attached graph ... (I made the colors match)
July 30th 2009, 05:45 PM
drkidd22

Thanks alot, that clears it up. May I ask what program/software you used to make the graph?
July 31st 2009, 05:39 AM
skeeter

Quote:

Originally Posted by drkidd22

Thanks alot, that clears it up. May I ask what program/software you used to make the graph?

Graph