Hi, I'm new to this forum and I need some help with this problem. If you could help, I'd appreciate it.
Find an equation of the tangent line at the point indicated.
f (x) = log2 (3x + x^-3), at x = 1
I tried using y - y1 = m(x - x1)
for y1, i got 2, x1 is 1, and for the slope i got (3-3/x^4)/((1/x^3+3 x) log(2))
so for the equation of the line, I got ((3-3/x^4)/((1/x^3+3 x) log(2)) (x-1)) +2
but unfortunately, its wrong.. lol