Question

Question #199ee

Answer 1

Write the function as:

`F(x) = sinx-2cosx = sqrt5(1/sqrt5sinx-2/sqrt5cosx)`

As `(1/sqrt5)^2+(2/sqrt5)^2 = 1`

we can put: `cos phi = 1/sqrt5` and `sin phi = 2/sqrt5`

so that:

`F(x) = sqrt5(cosphisinx-sinphicosx) = sqrt(5)cos(x+phi)`

with `tan phi = (2/sqrt5)/(1/sqrt5) = 2`, so `phi = arctan 2`

Thus the function is a sinusoid of amplitude `sqrt5` and phase `phi`. Accordingly it is decreasing when:

`0 < x < pi-phi`

increasing when:

`pi-phi < x < 2pi-phi`

and again decreasing for

`2pi-phi < x < 2pi`

It is concave up for:

`pi/2 -phi < x < (3pi)/2-phi`

and concave down in the rest of the interval.

graph{sqrt5cos(x+1.1071487178) [-0.1, 6.3, -3, 3]}