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I am not sure how to go about solving this problem:

enter image description here

I know the derivative of $f$ would be $2x$ but I am not sure where to go from there. If anyone could help out that would be great. Thanks!

Edit: The answer is: $y = 2x - 1$ , see below for why.

EDIT: For this problem:

enter image description here

plugging in $x_1$ back into the function did not seem to yield the correct $y_1$ to use in the point slope formula according to the solution software I use: enter image description here

I plugged in $x_1$ into the function $f(x)=3x^3$ to get 3 for $y_1$. So the final equation would be $y - 3 = 3(x-1)$ -> $y=3x$

I don't see any errors being made. what could be the issue?

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Good job, Matt! –  amWhy Jun 4 '11 at 1:25
    
Matt, the problem in your second edit is that $f(x)=x^3$, not $3x^3$. So plugging in $\pm 1$ indeed returns $\pm 1$. –  yunone Jun 4 '11 at 1:58
    
doy, i plugged it back into the derivative thanks! –  Matt Jun 4 '11 at 2:19
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1 Answer

up vote 2 down vote accepted

Here's a hint to get started. Rewrite the equation of the given line as $y=2x+1$. This line has slope $2$, so the line you're trying to find should also have slope $2$, in order to be parallel to the given line. Also, any line tangent to $f$ has slope $2x$. If you set these two slopes equal to each other, you solve for the $x$-coordinate of a line tangent to $f$ with slope $2$, which is then parallel to the given line, as desired.

Once you have that, you can find a point on your desired line. Since you already know the needed slope, you then have a slope and a point on the line, and you can use point-slope form to write your equation.

Please ask if anything is unclear.

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Ok, I didn't know about setting the derivative and the slope equal to find an x for a point on the line, thanks. we need to use the point slope formula (nipissingu.ca/calculus/tutorials/lineargifs/point-slope.gif) to get the equation of the line we are looking for right? So setting the derivative and the slope equal gave us $x_1$ = 1 to use in the formula but how do we get the $y_1$ for that formula? –  Matt Jun 4 '11 at 1:14
    
@Matt, the $x$-coordinate you find is of the point at the intersection of the desired line and $f$, so the point is also on $f(x)=x^2$. You then can just plug in $x_1$ to get $y_1$. –  yunone Jun 4 '11 at 1:16
    
solve for $y_1$ by substituting $x_1$ into the equation $y = x^2$, then you'll have the point ($x_1, y_1)$, together with the needed slope, and can thus find the equation of the line tangent to the parabola, parallel to the given line. (just saw your post, yunone; I'm pokey with my writing answers and comments; by the time I finish, someone has often already posted!) –  amWhy Jun 4 '11 at 1:18
    
Ah, ok, thanks so much! –  Matt Jun 4 '11 at 1:18
    
Sure, glad to help. –  yunone Jun 4 '11 at 1:19
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