# How do use the first derivative test to determine the local extrema (x^2-10x)^4?

Aug 28, 2015

The function has local (and global) minima at $x = 0$ and $x = 10$ and a local maximum at $x = 5$.

#### Explanation:

Let $f \left(x\right) = {\left({x}^{2} - 10 x\right)}^{4}$, then the Power Rule and Chain Rule give a derivative of

$f ' \left(x\right) = 4 {\left({x}^{2} - 10 x\right)}^{3} \cdot \left(2 x - 10\right)$

$= 8 {x}^{3} {\left(x - 10\right)}^{3} \left(x - 5\right)$

$f ' \left(x\right)$ is therefore clearly zero at $x = 0 , 5 , 10$, and these are the critical points of $f$.

A bit less clearly, $f '$ changes sign from negative to positive as $x$ increases through $x = 0$, from positive to negative as $x$ increases through $x = 5$, and from negative to positive as $x$ increases through $x = 10$. You can plug numbers less than 0, between 0 and 5, between 5 and 10, and greater than 10 and use the continuity of $f '$ to check this. You can also graph $f '$ to confirm it graphically.

The First Derivative Test then implies that the function has local (and global) minima at $x = 0$ and $x = 10$ and a local maximum at $x = 5$.

Interestingly, application of the Second Derivative Test in this example only allows us to conclude where the local maximum is, not where the local minima are (it turns out that $f ' ' \left(0\right) = f ' ' \left(10\right) = 0$ and $f ' ' \left(5\right) = - 125000 < 0$). In other words, the First Derivative Test is "stronger" (it can be used in more situations) than the Second Derivative Test.