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
$\displaystyle
f(x) = x^3
$
$\displaystyle f'(x) = 3x^2 $

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

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

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

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

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

My point is $\displaystyle (1,4)$

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

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

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

Quote:

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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
$\displaystyle
f(x) = x^3
$
$\displaystyle f'(x) = 3x^2 $

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

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

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

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

x = -1 also

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

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.

Quote:

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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
$\displaystyle
f(x) = x^3
$
$\displaystyle f'(x) = 3x^2 $

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

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

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

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

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

My point is $\displaystyle (1,5)$

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

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

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

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red part is wrong
if I understood the question you should enter the $\displaystyle x=1$ in the $\displaystyle
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

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

http://www.mathhelpforum.com/math-he...8f41dd91-1.gif this is the derrivative of $\displaystyle f(x)=x^3$ 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

Quote:

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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?

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

this says that there will be two distinct lines tangent to the curve y = x^3 that are parallel to $\displaystyle y = 3x+1$

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?

Quote:

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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?

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

$\displaystyle f'(x) = 3$

$\displaystyle 3x^2 = 3$

$\displaystyle x = \pm 1$

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

$\displaystyle y = x^3$

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

y - 1 = 3(x - 1)

y = 3x - 2 is the first line


$\displaystyle y = (-1)^3$ ... (-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)

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

Quote:

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Originally Posted by drkidd22

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

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Graph