What are the points of inflection, if any, of $f (x) = x^{6} + 3 x^{5} - (\frac{15}{2}) x^{4} - 40 x^{3} - 60 x^{2} + 8 x + 5$ $f(x) = x^6 + 3x^5 - (15/2)x^4 - 40x^3 - 60x^2 + 8x + 5$ ?

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Answer 1 · 2017-11-25T05:24:18

Points of inflection occur at $x = - 1, - 2$ $x=-1,-2$ and $2$ $2$ .

At point of inflection $\frac{d^{2} f (x)}{{d x}^{2}} = 0$ $(d^2f(x))/(dx^2)=0$

As $f (x) = x^{6} + 3 x^{5} - (\frac{15}{2}) x^{4} - 40 x^{3} - 60 x^{2} + 8 x + 5$ $f(x) = x^6+3x^5-(15/2)x^4-40x^3-60x^2+8x+5$

$\frac{d f}{d x} = 6 x^{5} + 15 x^{4} - 30 x^{3} - 120 x^{2} - 120 x + 8$ $(df)/(dx)=6x^5+15x^4-30x^3-120x^2-120x+8$

and $\frac{d^{2} f (x)}{{d x}^{2}} = 30 x^{4} + 60 x^{3} - 90 x^{2} - 240 x - 120$ $(d^2f(x))/(dx^2)=30x^4+60x^3-90x^2-240x-120$

Hence points of inflection are given by $30 x^{4} + 60 x^{3} - 90 x^{2} - 240 x - 120 = 0$ $30x^4+60x^3-90x^2-240x-120=0$

or $x^{4} + 2 x^{3} - 3 x^{2} - 8 x - 4 = 0$ $x^4+2x^3-3x^2-8x-4=0$

Using factor theorem we have zeros at $- 1, - 2$ $-1,-2$ and $2$ $2$

$(x - 2) (x + 2) {(x + 1)}^{2} = 0$ $(x-2)(x+2)(x+1)^2=0$

Hence, points of inflection occur at $x = - 1, - 2$ $x=-1,-2$ and $2$ $2$ .

Graph not drrawn to scale (shrunk vertically).
graph{x^6+3x^5-(15/2)x^4-40x^3-60x^2+8x+5 [-4, 4, -200, 200]}

What are the points of inflection, if any, of f(x) = x^6 + 3x^5 - (15/2)x^4 - 40x^3 - 60x^2 + 8x + 5 f(x)=x6+3x5−(152)x4−40x3−60x2+8x+5f(x) = x^6 + 3x^5 - (15/2)x^4 - 40x^3 - 60x^2 + 8x + 5 ?