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

Answer:

The correct option is B.

Step-by-step 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.

$\cos(x+4)=0$

$x+4=\frac{\pi}{2}+n\pi$          $[\because \cos x=0, then x=\frac{\pi}{2}+n\pi,n\in Z]$

Subtract 4 from both sides.

$x=\frac{\pi}{2}+n\pi-4$

For n=-1,

$x=\frac{\pi}{2}+(-1)\pi-4=-5.57079632679$

For n=0,

$x=\frac{\pi}{2}+(0)\pi-4=-2.42920367321$

For n=1,

$x=\frac{\pi}{2}+(1)\pi-4=0.712388980385\approx 0.712$

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