How to find the inflection points of (x^7/42)-(x^5/10)+(x^3/6)+(3x)+4? I know that it involves finding the zeros of the 2nd derivative and looking at concavity at those points, but I can't get to the zeros...

Question

anonymous · Answer

to get POI (points of infection) find where f" is equal to 0, you should use the equation feature so i can follow the equation better

anonymous · Answer

$$((x^7)\div42) -((x^5)\div10)+((x^3)\div6)+3x+4 $$

callisto · Answer

$$f(x) = \frac{x^7}{42}-\frac{x^5}{10}+\frac{x^3}{6}+3x+4$$
What is f'(x) ?

anonymous · Answer

$$x^6/7 -x^4/2 +x^2/2+3$$

anonymous · Answer

And then f double prim is $$x^5-2x^3+x$$

callisto · Answer

f'(x) is not right, f''(x) is right.

anonymous · Answer

*prime

callisto · Answer

So, we need to solve
$$f''(x)=0$$$$x^5-2x^3+x =0$$Factorize the left side, what do you get?

anonymous · Answer

x($$x^4-2x^2+1$$) ?

callisto · Answer

Factorize $x^4 -2x^2 +1$ too..

callisto · Answer

$$(a-b)^2 = a^2-2ab+b^2$$Take a = x^2 and b = 1

anonymous · Answer

OOOOOOOOOOOOOOOOOOOH RIIIIIIIIIIIIIIIIGGGGGGGGGGGGHHHHHHHHHTT

So it would be x^2 -1 ?

callisto · Answer

No. Instead, it should be $(x^2 -1)^2$. Though, you can still factorize $x^2-1$

anonymous · Answer

So, would the final factorized version be x((x-1)(x+1))^2 ?

callisto · Answer

Yup! :)

anonymous · Answer

THANK YOU!

callisto · Answer

Welcome :)