Question

equation of tangent line for f(x) = x squared tan (4 - 3x) at the point x=0

Answer 1

you got the derivative?

Answer 2

i got -3x squared sec squared (4-3x) + 2x tan(4-3x) ...but i'm a bit rusty so im not sure thats correct

Answer 3

yeah that looks good
\[f'(x)=-3x^2\sec^2(4-3x)+2x\tan(4-3x)\]

Answer 4

and so \(f'(0)=0\)

Answer 5

you line is therefore horizontal

Answer 6

many thanks!

Answer 7

yw