I am stumped on this question (super-nerd myself, working ahead in my textbook just for fun), but was hoping you could explain your answer and how you got it on this one.
Part A) the slope of f(x) at x=pi
Part B) write an equation for the tangent to f(x) at x=pi
Part C) write an equation for the normal line to f(x) at x=pi
Thank you for your assistance!
Perhaps it would have been helpful if you told me exactly what you're having problems with or the level at which you're studying. Anyway, for a function f(x) we can find its slope f'(x) (read f prime x) by differentiating the function.
Given f(x) = sin(x)
The slope f'(x) = cos(x)
At x = π, the slope of f(x) (which is the slope of the tangent line at x = π) then has a value of cos(π) = -1
As we know from coordinate geometry, we can write the equation of a line with a slope m which passes through a point (x1, y1) as;
(y - y1)/(x - x1) = m
m = -1
x1 = π
y1 is the value of f(x) at x1 i.e sin(π) = 0
The equation is then
(y - 0)/(x - π) = -1
y = π - x
Again from coordinate geometry, if two lines are perpendicular then the product of their slope is -1. Taking the slope of the normal line as k, we must have that
mk = -1
k = -1/m = -1/-1 = 1
The normal line also passes through the same point (π, 0) and so its equation is
(y - 0)/(x - π) = 1
y = x - π