Question

For `f(x)=xsin^2x` what is the equation of the tangent line at `x=(3pi)/2`?

Answer 1

y = x. See the tangent-inclusive Socratic graph.

Explanation:

`f=xsin^2x`

`f=3/2pi=4.7124`

at `x = 3/2pi=4.7124`, nearly.

So, the point of contact P is `(3/2pi, 3/2pi)=(4.7124, 4.7124)`.

`f'=x(sin^2x)'+sin^2x(x)'=sinx(2xcosx+sinx)`

`f'=1`, at `x =3/2pi`.

So, the equation to the tangent at P is

`y-3/2pi=x-3/2pi`, giving y = x

graph{(xsinx sinx-y)(y-x)=0 [-10, 10, -5, 5]}

Overall graph, symmetric about O. The tangent at `x = 3/2pi` ought

to touch the curve, again, at `x = -3/2pi`. Interestingly, It is a

common tangent for x = `+_`(odd multiple of `pi/2`).

Now, see the next graph for this touch-crests tangency

graph{(xsinx sinx-y)(y-x)=0 [-100 100 -50, 50]}

graph{(xsinx sinx-y)(y-x)((x-4.7)^2+(y-4.7)^2-.001)=0 [3, 6, 0, 6.5]}

Tangent-specific graph, for `x = pi/2`