Question 965179
{{{f(x)=3x^4-8x^3+6x^2+1}}}
{{{df/dx=12x^3-24x^2+12x=12x(x^2-2x+1)=12x(x-1)^2}}}
{{{d(df/dx)/dx=36x^2-48x+12=12(3x^2-4x+1)=12(x-1)(3x-1)}}}
Decreasing: ({{{-infinity}}},{{{0}}})
Increasing: ({{{0}}},{{{infinity}}})
Absolute min : ({{{0}}},{{{1}}})
Local min : ({{{1}}},{{{2}}})
Concave up:({{{-infinity}}},{{{0}}})U({{{1}}},{{{infinity}}})
Concave down: ({{{0}}},{{{1}}})
*[illustration v1.JPG].
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{{{y=( (x(x+1)(x-2))/((x^2-2)(2x+3)))^(1/3)}}}
{{{ln(y)=(1/3)ln( (x(x+1)(x-2))/((x^2-2)(2x+3)))}}}
{{{3*ln(y)=ln( x(x+1)(x-2))-ln((x^2-2)(2x+3))}}}
{{{3*ln(y)=ln(x)+ln(x+1)+ln(x-2)-ln(x^2-2)-ln(2x+3)}}}
{{{3*(dy/y)=dx/x+dx/(x+1)+dx/(x-2)-(2xdx)/(x^2-2)-dx/(2x+3)}}}
{{{3*(dy/y)=(1/x+1/(x+1)+1/(x-2)-(2x)/(x^2-2)-1/(2x+3))dx}}}
{{{dy/dx=(y/3)(1/x+1/(x+1)+1/(x-2)-(2x)/(x^2-2)-1/(2x+3))}}}
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{{{f(x)=(x^2+1)(x^2+2)^(1/3)}}}
{{{df/dx=(x^2+1)(1/3)(x^2+2)^(-2/3)(2x)+(x^2+2)^(1/3)(2x)}}}
{{{df/dx=((2x)/(3(x^2+2)^(2/3)))(x^2+1+3(x^2+2))}}}
{{{df/dx=((2x)/(3(x^2+2)^(2/3)))(x^2+1+3x^2+6)}}}
{{{df/dx=((2x)/(3(x^2+2)^(2/3)))(4x^2+7)}}}
{{{df/dx=((2x)(4x^2+7))/(3(x^2+2)^(2/3)))}}}
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{{{f(x)= (x+1)^4( x^2-1)^(-1/2)}}}
{{{df/dx=(x+1)^4(-1/2)(x^2-1)^(-3/2)(2x)+(x^2-1)^(-1/2)4(x+1)^3}}}
{{{df/dx=((x+1)^3/(x^2-1)^(-3/2))((x+1)(x)+4(x^2-1))}}}
{{{df/dx=((x+1)^3/(x^2-1)^(-3/2))(-x^2-x+4x^2-4)}}}
{{{df/dx=((x+1)^3/(x^2-1)^(-3/2))(3x^2-x-4)}}}
{{{df/dx=((x+1)^3/(x^2-1)^(-3/2))((3x-4)(x+1))}}}
{{{df/dx=((x+1)^4(3x-4))/(x^2-1)^(-3/2)}}}