Question

How do you sketch the graph `y=sinx+sin^2x` using the first and second derivatives from `0<=x<2pi`?

Answer 1

Please see below.

Explanation:

We have `y=sinx+sin^2x=sinx(1+sinx)`

Observe that `y=0` when `sinx=0` or `1+sinx=0`

hence between `0 <= x <= 2pi`, it intersects `x`-axis at `x={0,pi,(3pi)/2,2pi}` or `{0,3.1416,4.7124,6.2832}`

Now we have extrema when `(dy)/(dx)` and as

`(dy)/(dx)=cosx+2sinxcosx=cosx(1+2sinx)` or `cosx+sin2x`

also `(d^2y)/(dx^2)=-sinx+2cos2x`

Hence extrema occurs when `cosx=0` i.e. `{pi/2,(3pi)/2}`. While second derivative is negative at both these points, when `1+2sinx=0` i.e. `sinx=-1/2` or `x={(5pi)/3,(7pi)/3}`, it is positive.

Hence we have a local minima at `x={(5pi)/3,(7pi)/3}` and local maxima at `{pi/2,(3pi)/2}`.

The graph appears as shown below.

graph{sinx(1+sinx) [-2.917, 7.083, -1.84, 3.16]}