Question

How do you find all points on the graph of the function `f(x)=2sinx+sin^2x` at which the tangent line is horizontal?

Answer 1

The points are `x = pi/2 + 2pin and (3pi)/2 + 2pin`.

Explanation:

We know that the derivative represents the instantaneous rate of change of a function, or the slope of the function. We also know that the slope of a line that is horizontal is `0`. We start by differentiating the function.

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

`f(x) = 2sinx + (sinx)(sinx)`

`f'(x) = (0 xx sinx + 2 xx cosx) + (cosxsinx + cosxsinx)`

`f'(x) = 2cosx + 2cosxsinx`

We set this to `0` and solve for `x`.

`0 = 2cosx + 2cosxsinx`

`0 = 2(cosx + cosxsinx)`

`0 = cosx(1 + sinx)`

`cosx = 0 and sinx= -1`

`x = pi/2 + 2pin, (3pi)/2 + 2pin`

Hopefully this helps!