find all points on the graph of f at which the tangent line is horizontal

Question

a) f(x)=(x+1)/(x^2+3)
b) y=cos(x)+sin(x),0≤x≤2π

Answer 1

For the tangent line to be horizontal, the slope must be zero
f(x)=(x+1)/(x^2+3)
f'(x) = ( (x^2 + 3)(1) - (x+1)(2x) )/(x^2 + 3)^2
= 0

x^2 + 3 - 2x^2 - 2x = 0
-x^2 - 2x + 3 = 0
x^2 + 2x - 3 = 0
(x + 3)(x - 1) = 0
x = -3 , x = 1

if x = -3, f(x) = (-3+1)/(9 + 3) = -2/12 = -1/6
so one point is (-3,-1/6)

if x = 1 ......... (your turn)

y = cosx + sinx
dy/dx = -sinx + cosx = 0 for horizonatal tangent
sinx = cosx
sinx/cosx = 1
tanx = 1
x = π/4, 5π/4, (45°, 225°) ,

(9π/4 would be the next one, but that's outside domain)

if x = π/4
y = sinπ/4 + cosπ/4 = √2/2 + √2/2 = √2
so one point is (π/4, √2)

your turn to find the other point

Answer 2

or, note that cosx + sinx = √2 sin(x + π/4)
so the tangent is horizontal when
cos(x + π/4) = 0
That is when x + π/4 = π/2 or 3π/2
one solution, given above, is x = π/4
your turn to find the other