f(x) = tan2(x) + (√3 - 1)[tan(x)] - √3 = 0
tan2(x) + √3[tan(x)] - tan(x) - √3 = 0
Factor into
[-1 + tan(x)]*[√3 + tan(x)] = 0
which means
[-1 + tan(x)] = 0 and/or [√3 + tan(x)] = 0
Then
tan(x) = 1
tan-1(1) = pi/4 radians
For the other equation
[√3 + tan(x)] = 0
tan(x) = -√3
tan-1(-√3) = -pi/3
so that
x = pi/4 or -pi/3 in the interval [0, 2pi]
Answer:

Step-by-step explanation:
From the question we are told that:
Function

Generally 1st derivative f'(x) of Fuction f(x) is


Generally 2nd derivative f''(x) of Fuction f(x) is
Where

Therefore



The slope-intercept form: y = mx + b
m - a slope
b - a y-intercept
y = x + 3
Answer: the slope m = 1, y-intercept b = 3 -> (0; 3).
1 foot, 8 inches is the correct answer
I wrote my decision on a piece of paper.