Let the required equation be y = mx + c; where y = 1, m = f'(0), x = 0
f(a) = sec(a)
f'(a) = sec(a)tan(a)
f'(0) = sec(0)tan(0) = 0
y = mx + c
1 = 0(0) + c
c = 1
Therefore, required equation is
y = 1.
I would say y = ln x + 4
you can use a graphing calculator
Answer: The real part is -6
The imaginary part is 2i
Step-by-step explanation:
