identify the coordinates of any local and absolute extreme points and inflection points. Graph the function.

Question

y = √1-x²÷2x+1

Answer 1

local min (1,0)

local max (-1,0)

no global max or min, y approaches infinity

inflection points (1/2, 1/2), (-3/4, 3/4)

try graphing

(sqr(1-x^2)/(2x+1)

local max & min are x intercepts

but if you really meant sqr1 -x^2/2x +1 = 2-x/2

then there are no max, min or inflection points

it's just a downward sloping straight line

sqr1 -x^2/2x + 1 as written means sqr1 -(x^2/2x)+1 =1 -x/2+1 = 2-x/2