# What are the points of inflection, if any, of f(x)=x^7/42 - (3x^6)/10 + (6x^5)/5 - (4x^4)/3 ?

Dec 9, 2015

The only inflection point is $\left(1 , - \frac{43}{105}\right)$

#### Explanation:

$f \left(x\right) = {x}^{7} / 42 - \frac{3 {x}^{6}}{10} + \frac{6 {x}^{5}}{5} - \frac{4 {x}^{4}}{3}$

$f ' \left(x\right) = {x}^{6} / 6 - \frac{9 {x}^{5}}{5} + 6 {x}^{4} - \frac{16 {x}^{3}}{3}$

$f ' ' \left(x\right) = {x}^{5} - 9 {x}^{4} + 24 {x}^{3} - 16 {x}^{2}$

In order to determine where the concavity does change, we must first determine where it might change.

$f ' ' \left(x\right)$ is never undefined, so we need to solve $f ' ' \left(x\right) = -$

$f ' ' \left(x\right) = {x}^{2} \left({x}^{3} - 9 {x}^{2} + 24 x - 16\right)$

Use inspection or the rational zeros test to see that $1$ is a zero of ${x}^{3} - 9 {x}^{2} + 24 x - 16$.

Then use division or trial and error to factor

${x}^{3} - 9 {x}^{2} + 24 x - 16 = \left(x - 1\right) \left({x}^{2} - 8 x + 16\right)$

$= \left(x - 1\right) {\left(x - 4\right)}^{2}$

This gives us:

$f ' ' \left(x\right) = {x}^{2} \left(x - 1\right) {\left(x - 4\right)}^{2}$. So,

$f ' ' \left(x\right) = 0$ at $0 , 1 , \text{and } 4$.

But $0$ and $4$ are zeros of even multiplicity, so the sign of $f ' ' \left(x\right)$ does not change at those zeros.

The sign of $f ' ' \left(x\right)$ changes only at $x = 1$.

( Note: this could also be determined by using a sign chart for $f ' '$.)

The inflection pint is the point $\left(1 , f \left(1\right)\right)$, so we do the arithmetic to find $f \left(1\right) = - \frac{43}{105}$, and

The inflection point is $\left(1 , - \frac{43}{105}\right)$