Therefore, the inflection points are at x=-1/\sqrt{2} and x=1/\sqrt{2}.

[wildcat12]

Homework Statement

find the inflection points of the curve y=ln(x^3-3x+4) correct to six decimal places

Homework Equations

The Attempt at a Solution

my first derivative was (3x^2)-3/(x^3)-3x+4) and my 2nd derivative as (-3x^3)+24x+9/(x^3)-3x+4). i do not know how to set this equal to zero

 

[vela]

First off, you need to use parentheses appropriately to avoid confusion. As written, it looks like you're saying the first derivative is equal to

$$y'=3x^2 - \frac{3}{x^3-3x+4}$$

whereas what you meant was

$$y'=\frac{3x^2-3}{x^3-3x+4}$$

Your second derivative looks wrong to me. You should recheck your calculations.

To set the second derivative equal to 0, you just write "=0" after your result. Solve for x as usual.

 

[amin001]

if it is y=x^4-3x^2+2.
How to find the inflection point??

 

[hunt_mat]

$$\frac{d^{2}y}{dx^{2}}=12x^{2}-6$$
Set this equal to zero to find that $$x=\pm 1/\sqrt{2}$$

 

[gb7nash]

$$\frac{d^{2}y}{dx^{2}}=12x^{2}-6$$
Set this equal to zero to find that $$x=\pm 1/\sqrt{2}$$

Note that this does not automatically make these inflection points. You still need to test points around +-1/sqrt(2).

If the sign changes from + to - or vice versa, then you have an inflection point.