# How do you test f(x)=8 x^4−9 x^3 +9 for concavity and inflection points?

Apr 11, 2018

#### Explanation:

To test for the concavity and inflection points you need to equate the second order derivative with zero.

Keeping in mind:

• $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$

• $\frac{d}{\mathrm{dx}} c = 0$

We proceed:

$f \left(x\right) = 8 {x}^{4} - 9 {x}^{3} + 9$

$\implies f ' \left(x\right) = 32 {x}^{3} - 27 {x}^{2}$

$\implies f ' ' \left(x\right) = 96 {x}^{2} - 54 x$

$= 6 \left(16 {x}^{2} - 9 x\right)$

Now,

$f ' ' \left(x\right) = 0$

$\implies 6 \left(16 {x}^{2} - 9 x\right) = 0$

$\implies x \left(16 x - 9\right) = 0$

$\implies \textcolor{red}{x = 0 , x = \frac{9}{16}}$ are the inflection points. Inflection points are those points where the curve changes its concavity if any.

graph{x(16x -9) [-5, 5, -5, 5]}

Sign Chart: See image.

Now, to determine the opening of the concavity.

• Put any value less than $0$ in $f ' ' \left(x\right)$.

$f ' ' \left(x\right)$ comes out to be positive. $\left(+\right)$.

• Put any value between $0$ to $\frac{9}{16}$.

$f ' ' \left(x\right)$ comes out to be negative. $\left(-\right)$.

• Put any value greater than $\frac{9}{16}$.

$f ' ' \left(x\right)$ comes out to be positive. $\left(+\right)$.

Thus, (-oo,0)∪(9/(16),oo) our concavity is upwards.
$\left(0 , \frac{9}{16}\right)$ our concavity is downwards.