Question #199ee

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Answer 1 · 2017-08-07T22:40:24

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]}