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

Jul 30, 2017

The points of inflection are:
$\left(- 3.2770 , 849.8146\right) \mathmr{and} \left(0.6103 , 0.3033\right)$

#### Explanation:

Points of inflection are points on the function where the second derivative changes sign, that is the graph goes from concave upward to concave downwards or vice-versa.

Let $y = f \left(x\right)$

We must find the second derivative of the function:
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = y ' ' \left(x\right)$

$y ' \left(x\right) = - 6 {x}^{5} - 20 {x}^{4} + 20 {x}^{3}$, using the power rule

Then,

$y ' ' \left(x\right) = - 30 {x}^{4} - 80 {x}^{3} + 60 {x}^{2}$

Let us find where the second derivative is zero, to find the critical x-values.

$- 30 {x}^{4} - 80 {x}^{3} + 60 {x}^{2} = 0$
$- 10 {x}^{2} \left(3 {x}^{2} + 8 x - 6\right) = 0$

$x = 0$ is a solution here, and to find the other two solutions, we solve the following equation:

$3 {x}^{2} + 8 x - 6 = 0$

x = (-8 ± sqrt(64 - (-72)))/(6) = (-8 ± sqrt(136))/(6)
$x \approx - 3.27698396495 , \mathmr{and} x \approx 0.610317298282$

Let us find the concavity of $f \left(x\right)$ on these intervals:

$o n \left(- \infty , - 3.27698396495\right) , f ' ' \left(x\right) < 0$
$o n \left(- 3.27698396495 , 0\right) , f ' ' \left(x\right) > 0$
$o n \left(0 , 0.610317298282\right) , f ' ' \left(x\right) > 0$
$o n \left(0.610317298282 , \infty\right) , f ' ' \left(x\right) < 0$

So in succession on $\left(- \infty , \infty\right)$, the sign line looks like this:
$- , + , + , -$

Therefore, $f ' ' \left(x\right)$ only changes signs at the x-values of $- 3.27698396495$ and $0.610317298282$

Since $f ' ' \left(x\right)$ changes signs at these values, $f \left(x\right)$ has inflection points here as well. The coordinates of the inflection points are:

$\left(- 3.27698396495 , f \left(- 3.27698396495\right)\right)$

and

$\left(0.610317298282 , f \left(0.610317298282\right)\right)$

Therefore, the coordinates are, rounding to 4 decimal places:

$\left(- 3.2770 , 849.8146\right) \mathmr{and} \left(0.6103 , 0.3033\right)$