Hi!
Is my solution correct?
Let . Find the equation of the line tangent to the graph of at the point where .
Limit at doesn't exist.
Still not going well. I don't think I've encountered that form for a line, or maybe I'm just not seeing it right. Try the slope-intercept form of a line:
y = mx + b
The slope for x = 2 is 2x = 2(2) = 4. So we have
y = 4x + b
You need to have the actual point where the tangent occurs, so what is f(2)? That will give you your point: (x, y) = (2, f(2)), and you plug that into your tangent line equation. That will give you the intercept for your tangent line.
By the way, recalculate f(2). It's not 4.
-Dan