f(x) = 3x + 2.cos(x) + 5 ← this is a function

f'(x) = 3 - 2.sin(x) ← this is its derivative

Then you solve for x the equatio: f'(x) = 0

3 - 2.sin(x) = 0

2.sin(x) = 3

sin(x) = 3/2 ← no possible

What the sign of f'(x) is ? → you know that: - 1 < sin(x) < 1

- 1 < sin(x) < 1 → you multiply by (- 1), don't forget to modify the direction

1 > - sin(x) > - 1 → then you multiply by (2)

2 > - 2.sin(x) > - 2 → and you add 3

2 + 3 > - 2.sin(x) + 3 > - 2 + 3 → you can simplify

5 > 3 - 2.sin(x) + 3 > 1 → recall: f'(x) = 3 - 2.sin(x)

5 > f'(x) > 1

So, you can see that f'(x) belongs to the interval [1 ; 5].

So you can say that: f'(x) > 0

Then you can conclude that the function is always increasing.

As the function sis always increasing, the representative curve of the function crosses only once the x-axis.

f(x) = 0 ← there is only one solution

Newton's Method

https://www.youtube.com/watch?v=h0XUxE47l00

Assuming: x = - π/2

f(- π/2) = 3.(- π/2) + 2.cos(- π/2) + 5

f(- π/2) = (- 3π/2) + 5 = (10 - 3π)/2

→ point A [- π/2 ; (10 - 3π)/2]

f'(- π/2) = 3 - 2.sin(- π/2)

f'(- π/2) = 3 + 2

f'(- π/2) = 5 ← this is the slope of the tangent line to the curve at: x = - π/2

According to the method, you can write:

x₀ = (- π/2) - { [(10 - 3π)/2] / 5 }

x₀ = (- π/2) - [(10 - 3π)/10]

x₀ = [- 5π - (10 - 3π)]/10

x₀ = [- 5π - 10 + 3π]/10

x₀ = (- 10 - 2π)/10

x₀ = (- 5 - π)/5

x₀ ≈ - 1.62831853

the derivative , 3 - 2 sin x , is > 0 ===> only one root......on a TI do the following , - 3 , enter, type in ( 5 + 2 cos ans + 2 ans sin ans ) / ( 2 sin ans - 3 ) , enter, enter,enter

to find ≈ - 1.62833