How do you find the inflection points of f(x)=3x^5-5x^4-40x^3+120x^2?

Jun 7, 2017

The inflection points are $x = - 2 , 1 , 2$

Explanation:

To find the inflection points you need to perform the second derivative test. Since this is a polynomial we use the power rule to differentiate the equation, $n {x}^{n - 1}$.
We get the first $\frac{d}{\mathrm{dx}}$ which is:

$f ' = 15 {x}^{4} - 20 {x}^{3} - 120 {x}^{2} + 240 x$

Followed by the second $\frac{d}{\mathrm{dx}}$ which is:

$f ' ' = 60 {x}^{3} - 60 {x}^{2} - 240 x + 240$

Now we factor you should get:

$60 {x}^{2} \left(x - 1\right) - 240 \left(x - 1\right)$

$\left(60 {x}^{2} - 240\right) \left(x - 1\right)$

Now set the factors equal to zero:

$60 {x}^{2} - 240 = 0$ and $x - 1 = 0$

Solve them and you should get:

$x = \pm 2 , 1$

If you wish to find out where the exact inflection points occur plug in the three values into the original equation :)
You would do $f \left(- 2\right) , f \left(1\right) , f \left(2\right)$.