# Answer to Question #17676 in Calculus for hsd

Question #17676

Let f(x)=1/(3x^(2)+1). Find the open intervals on which f is concave up (down). Then determine the x-coordinates of all inflection points of f.

Use interval notation if you are asked to find an interval or union of intervals. (This link opens instructions below)

f is concave up on the intervals

f is concave down on the intervals

The inflection points occur at x =

Use interval notation if you are asked to find an interval or union of intervals. (This link opens instructions below)

f is concave up on the intervals

f is concave down on the intervals

The inflection points occur at x =

Expert's answer

find inflection points we must find where f''=0

f'=-6x/(3x^2+1)^2

f''=(6 (-1+9 x^2))/(1+3 x^2)^3

f"=0 9x^2=1

x=1/3 or x=-1/3

inflection points occur at x =1/3 and x=-1/3to define is function concave up or down we must checkf"<0 or f">0 when

x<-1/3 then f">0 then f concave down -infinity<x<-1/3-1/3<x<1/3 f"<0 then f concave up

x>1/3 f"<0 then f concave down

f'=-6x/(3x^2+1)^2

f''=(6 (-1+9 x^2))/(1+3 x^2)^3

f"=0 9x^2=1

x=1/3 or x=-1/3

inflection points occur at x =1/3 and x=-1/3to define is function concave up or down we must checkf"<0 or f">0 when

x<-1/3 then f">0 then f concave down -infinity<x<-1/3-1/3<x<1/3 f"<0 then f concave up

x>1/3 f"<0 then f concave down

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