Determine the values at which the inflection points occur, if any, of the given function. 𝑓(𝑥)=𝑥/𝑥^2+25

Answer 1

f(x) = x/x^2 +25 = 1/x +25 is a rectangular hyperbola with no inflection points. It has 2 branches, both everywhere downward sloping, always decreasing, with a discontinuity at x=0

but f(x) = x/(x^2 +25) is a different graph. You probably intended to write this, but the parentheses are missing.

take the derivative and set = zero

f'(x) = -2x^2/(x^2 + 25)^2 + 1/(x^2 +25) =0

(-2x^2 +x^2+25)/(x^2+25)^2 = 0

(25-x^2)/(x^2+25)^2 = 0

set the numerator = 0

25-x^2 = 0

x^2 = 25

x = 5 or -5

f(5) = 10

f(-5) = -10

critical points are (5,1/10) and (-5,-1/10), but neither is an inflection point as neither f'(5) or f"(-5) = 0

(5,10) is a local maximum, (-5,-10) is a local minimum

take the 2nd derivative, to determine local max, local min and any inflection point

the derivative calculations get complicated, so an error is easy to make

f"(x) = (2x^3-150x)/(x^2+25)^3

set numerator = 0, solve for x

2x(x^2 -75) = 0, x= 0, x = 5sqr3, x=-5sqr3

f(0) =0, (0,0), the origin is an inflection point

f(5sqr3) = 5sqr3/(75+25) = sqr3/20, (5sqr3, .05sqr3) is another infleciton point

f(-5sqr3) = -5sqr3(75+25) = -sqr3/20 (-5sqr3, -.05sqr3) is a 3rd inflection point