# How do you find the maximum, minimum and inflection points and concavity for the function g(x) = 170 + 8x^3 + x^4?

Point of minima $x = - 6$, points of inflection $x = 0 , x = - 4$ & the function is concave in $x \setminus \in \left(- 6 , 0\right)$

#### Explanation:

The given function:

$g \left(x\right) = 170 + 8 {x}^{3} + {x}^{4}$

$g ' \left(x\right) = 24 {x}^{2} + 4 {x}^{3}$

$g ' ' \left(x\right) = 48 x + 12 {x}^{2}$

For mamima or minima we have

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

$24 {x}^{2} + 4 {x}^{3} = 0$

$4 {x}^{2} \left(6 + x\right) = 0$

$x = 0 , - 6$

Now, $g ' ' \left(0\right) = 48 \left(0\right) + 12 {\left(0\right)}^{2} = 0$

hence, $x = 0$ is a point of inflection

Now, $g ' ' \left(- 6\right) = 48 \left(- 6\right) + 12 {\left(- 6\right)}^{2}$

$= 144 > 0$

hence, $x = - 6$ is a point of maxima

Now, for points of inflection we have

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

$48 x + 12 {x}^{2} = 0$

$12 x \left(4 + x\right) = 0$

$x = 0 , - 4$

The curve will be concave iff

$g ' \left(x\right) < 0$

$24 {x}^{2} + 4 {x}^{3} < 0$

$4 {x}^{2} \left(6 + x\right) < 0$

$x \setminus \in \left(- 6 , 0\right)$

Thus, the graph will be concave for $x \setminus \in \left(- 6 , 0\right)$