# How do you use the first derivative test to determine the intervals on which f(x)=x^4+3x^3+3x^2+1 is increasing or decreasing and whether each critical point is a local maximum, minimum or neither?

Sep 24, 2015

See the explanation.

#### Explanation:

Using the rules:
$y = {x}^{n} , y ' = n {x}^{n - 1}$

and

$h \left(x\right) = {f}_{1} \left(x\right) + {f}_{2} \left(x\right) + \ldots + {f}_{n} \left(x\right)$
$h ' \left(x\right) = f {'}_{1} \left(x\right) + f {'}_{2} \left(x\right) + \ldots + f {'}_{n} \left(x\right)$

we get:

$f ' \left(x\right) = 4 {x}^{3} + 9 {x}^{2} + 6 x = x \left(4 {x}^{2} + 9 x + 6\right)$

Lets examine expression $4 {x}^{2} + 9 x + 6$:
$4 {x}^{2} + 9 x + 6 = 0$
$D = {b}^{2} - 4 a c = {9}^{2} - 4 \cdot 4 \cdot 6 = 81 - 96 = - 15$
$D < 0$ which means that quadratic function $4 {x}^{2} + 9 x + 6$ doesn't have real zeros. Furthermore, $a = 4 > 0$ (coefficient in front of the x^2), so the function is concave.

So, expression $4 {x}^{2} + 9 x + 6 > 0$ for $\forall x \in R$.

$f ' \left(x\right) = 0 \iff x \left(4 {x}^{2} + 9 x + 6\right) = 0 \iff x = 0$

$\forall x < 0 : f ' \left(x\right) < 0$ and $f$ is decreasing
$\forall x > 0 : f ' \left(x\right) > 0$ and $f$ is increasing

$f ' \left(x\right)$ changes sign in $x = 0$ and $f \left(x\right)$ has minimum value ${f}_{\min} = f \left(0\right) = 1$