Find all lines that are tangent to the curve y = x^3 and are also parallel to the line 3x - y = 5
Any help will be appreciated!
Firstly,
$\displaystyle y=x^3$
$\displaystyle y'=3x^2$
The lines must be parallel to 3x-y = 5, $\displaystyle y=3x-5$. So their slope must be $\displaystyle m=3$.
Let our line be $\displaystyle y=m\cdot x + C$, m is the slope and C is a constant. As $\displaystyle m=3$ it'll become $\displaystyle y=3x + C$
At the tangent point, the line must satisfy $\displaystyle m=3=3x^2$.
$\displaystyle 3x^2 = 3$
$\displaystyle x^2=1$
$\displaystyle x= \pm 1$
So the line passes the points (-1,f(-1)) and (1,f(1)).
(-1,-1), (1,1)
The line is $\displaystyle y=3x + C$
And it makes $\displaystyle C=2$ for (-1,1)
and $\displaystyle C=-2$ for (-1,1)
The lines are then
$\displaystyle y=3x + 2 $ and $\displaystyle y=3x - 2$
bu arada, türk müsün? =)