# How many inflection points are in the graph of f(x)= (x^7)/42 - (3x^6)/10 + (6x^5)/5 - (4x^4)/3?

Aug 1, 2015

There is one inflection point.

#### Explanation:

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

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

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

Finding the zeros of $f ' '$

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

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

Rational Zeros Theorem and checking (or inspection of likely candidates or adding the coefficients) we see that $1$ is a zero. Therefore $x - 1$ is a factor.
Factor or do the division to get:

$f ' ' \left(x\right) = {x}^{2} \left(x - 1\right) \left({x}^{2} - 8 x + 16\right)$

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

The zeros are $0 , 1 , \text{and } 4$.

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

Therefore the only inflection point is $\left(1 , f \left(1\right)\right)$.