Question

To three decimal places, find the value of the first positive x-intercept for the function f(x) = 2cos(x + 4)

1.712
0.712
–2.429
–2.712

Answer 1

The correct answer would be B) 0.712

Answer 2

The correct option is B.

Explanation:

The given function is

`f(x)=2\cos(x+4)`

We need to find the first positive x-intercept for the function f(x).

Equate f(x)=0, to find the x-intercepts.

`f(x)=0`

`2\cos(x+4)=0`

Divide both sides by 2.

`\cos(x+4)=0`

`x+4=(\pi)/(2)+n\pi`
`[\because \cos x=0, then x=(\pi)/(2)+n\pi,n\in Z]`

Subtract 4 from both sides.

`x=(\pi)/(2)+n\pi-4`

For n=-1,

`x=(\pi)/(2)+(-1)\pi-4=-5.57079632679`

For n=0,

`x=(\pi)/(2)+(0)\pi-4=-2.42920367321`

For n=1,

`x=(\pi)/(2)+(1)\pi-4=0.712388980385\approx 0.712`

The first positive x-intercept for the function f(x) is 0.712. Therefore the correct option is B.