# What does the 2nd Derivative Test tell you about the behavior of f(x) = x^4(x-1)^3 at these critical numbers?

Aug 10, 2015

The Second Derivative Test implies that the critical number (point) $x = \frac{4}{7}$ gives a local minimum for $f$ while saying nothing about the nature of $f$ at the critical numbers (points) $x = 0 , 1$.

#### Explanation:

If $f \left(x\right) = {x}^{4} {\left(x - 1\right)}^{3}$, then the Product Rule says

$f ' \left(x\right) = 4 {x}^{3} {\left(x - 1\right)}^{3} + {x}^{4} \cdot 3 {\left(x - 1\right)}^{2}$

$= {x}^{3} \cdot {\left(x - 1\right)}^{2} \cdot \left(4 \left(x - 1\right) + 3 x\right)$

$= {x}^{3} \cdot {\left(x - 1\right)}^{2} \cdot \left(7 x - 4\right)$

Setting this equal to zero and solving for $x$ implies that $f$ has critical numbers (points) at $x = 0 , \frac{4}{7} , 1$.

Using the Product Rule again gives:

$f ' ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{3} \cdot {\left(x - 1\right)}^{2}\right) \cdot \left(7 x - 4\right) + {x}^{3} \cdot {\left(x - 1\right)}^{2} \cdot 7$

$= \left(3 {x}^{2} \cdot {\left(x - 1\right)}^{2} + {x}^{3} \cdot 2 \left(x - 1\right)\right) \cdot \left(7 x - 4\right) + 7 {x}^{3} \cdot {\left(x - 1\right)}^{2}$

$= {x}^{2} \cdot \left(x - 1\right) \cdot \left(\left(3 x - 3 + 2 x\right) \cdot \left(7 x - 4\right) + 7 {x}^{2} - 7 x\right)$

$= {x}^{2} \cdot \left(x - 1\right) \cdot \left(42 {x}^{2} - 48 x + 12\right)$

$= 6 {x}^{2} \cdot \left(x - 1\right) \cdot \left(7 {x}^{2} - 8 x + 2\right)$

Now $f ' ' \left(0\right) = 0$, $f ' ' \left(1\right) = 0$, and $f ' ' \left(\frac{4}{7}\right) = \frac{576}{2401} > 0$.

The Second Derivative Test therefore implies that the critical number (point) $x = \frac{4}{7}$ gives a local minimum for $f$ while saying nothing about the nature of $f$ at the critical numbers (points) $x = 0 , 1$.

In actuality, the critical number (point) at $x = 0$ gives a local maximum for $f$ (and the First Derivative Test is strong enough to imply this, even though the Second Derivative Test gave no information) and the critical number (point) at $x = 1$ gives neither a local max nor min for $f$, but a (one-dimensional) "saddle point".