1
$\begingroup$

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?

  • 0
    doy, i plugged it back into the derivative thanks!2011-06-04

1 Answers 1

2

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.

  • 0
    I think I need to dig a bit deeper with this concept. I am still having trouble. Please see my 2nd edit if you are willing to assist me!2011-06-04