# What are the points of inflection, if any, of f(x)= x/sqrt(x^5-x^3-3x+4) ?

Aug 30, 2017

There are 3 inflection points at x≈-0.797033, x≈0.914124 and x≈1.36922

To get the y-coordinates, just substitute the above x-values into $f \left(x\right)$

#### Explanation:

The points of inflection are the roots of the second derivative, $f ' ' \left(x\right) = 0$, which is $f ' ' \left(x\right) = \frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = x \left(\frac{3 {\left(5 {x}^{4} - 3 {x}^{2} - 3\right)}^{2}}{4 {\left({x}^{5} - {x}^{3} - 3 x + 4\right)}^{\frac{5}{2}}} - \frac{20 {x}^{3} - 6 x}{2 {\left({x}^{5} - {x}^{3} - 3 x + 4\right)}^{\frac{3}{2}}}\right) - \frac{5 {x}^{4} - 3 {x}^{2} - 3}{{x}^{5} - {x}^{3} - 3 x + 4} ^ \left(\frac{3}{2}\right)$ obtained by diffentiating the derivative of $f \left(x\right)$

You would get 3 real solutions, thus 3 inflection points, at x≈-0.797033, x≈0.914124 and x≈1.36922

You can verify visually that those are the inflection points from the graph
graph{x/sqrt(x^5-x^3-3x+4) [-10, 10, -5, 5]}

Here is an explanation of inflection points by Sal Khan from Khan Academy

Summary

Essentially, inflection points are where a curve is changing concavity, where the function is neither concave up nor concave down, but rather transitioning from one to the other.

Mathematically, when a curve is concave up, the second derivative of the curve is negative (as the slope is decreasing) and when the curve is concave downwards, the second derivative is positive.

Thus, when the function is at an inflection point, the second derivative is neither positive nor negative, and thus equals to 0