Find all points on the graph of f(x)=xe^x at which the tangent line is parallel to the line y-3x=4.
i figured out that f'(x)=3 but i don't know what to do with that.
Indeed, you need to equate f'(x)=3 and solve for x.
First find the slope of the line
L: y-3x=4
L: y=3x+4
slope = 3
To find f'(x), we start with f(x) and do implicit differentiation:
f(x)=y=xe^x
take ln
ln(y)=ln(x)+xln(e)
ln(y)=x+ln(x)
differentiate with respect to x:
y'/y = 1+1/x
y' = (1+1/x)y
=(1+1/x)xe^x
=(1+x)e^x
So solve for y' = slope = 3
I get x=0.6 approx.
There should be only one root because the function y' is strictly increasing.