6. Given the function 𝑝(𝑥) = (𝑚𝑥)/(𝑛 + 𝑞𝑥^2)

Question

M=1, N=2, Q=3,

Find (a) all critical points

(b) all extreme values

(c) all inflection points (give an ordered pair)

(d) identify a suitable interval that includes all these features, and

(e) provide a graph on this interval (not hand drawn; use desmos). Be sure to choose the graphing window appropriately so that the graph clearly shows the important features and there are no huge areas of empty space.

Answer 1

p(x) = x/(2+3x^2) = x/(3x^2+2) = x(3x^2+2)^(-1)

first derivative.....

quotient rule says:

p' = [(3x^2+2) - x(6x)] / (3x^2+2)^2

= [3x^2 +2 - 6x^2]/(3x^2+2)^2

= (-3x^2+2)/(3x^2+2)^2 = (-3x^2 + 2)(3x^2+2)^(-2)

product rule says:

-x[(3x^2+2)^(-2)](6x) + (3x^2+2)^(-1)

[3x^2+2)^(-2)] [(-6x^2) + 3x^2+2 ]

= (-3x^2+2)/(3x^2+2)^2

so the first derivative is correct....

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second derivative....

quotient rule says:

p'' = [(3x^2+2)^2 ( -6x) - 2(-3x^2+2)(3x^2+2)(6x)]/(3x^2+2)^4

= (3x^2+2) [(3x^2+2)(-6x) - 2(-3x^2+2)(6x)]/(3x^2+2)^4

= [(3x^2+2)(-6x) - 2(-3x^2+2)(6x)]/(3x^2+2)^3

= [ -18x^3 - 12x + 36x^3-24x]/(3x^2+2)^3

= [ 18x^3 - 36x]/(3x^2+2)^3

= (18x)(x^2-2)/(3x^2+2)^3

product rule says:

(-3x^2+2)(-2)(3x^2+2)^(-3) (6x) + (-6x)(3x^2+2)^(-2) =

(3x^2+2)^(-3) [ (-3x^2+2)(-12x) + (-6x)(3x^2+2) ] =

(3x^2+2)^(-3) [ 36x^3 - 24x - 18x^3 - 12x ] =

(3x^2+2)^(-3) [ 18x^3 - 36x] =

(18x)(x^2 - 2) / (3x^2 +2 )^(3)

so the second derivative is correct

(0,0) is the only solution...

setting to zero , the first derivative has solutions x^2 = 2/3 ---> x = +or- sqrt(6)/3

critical points (+or- sqrt(6)/3, +or-sqrt(6)/12)

setting to zero, the second derivaive has the same solutions , PLUS (0,0);

so the graph has flex point at origin