# Question #041ca

Oct 20, 2017

The function is increasing on the intervals $\left(- \infty , - 4\right]$ and $\left[4 , \infty\right)$ and is decreasing on the intervals $\left[- 4 , 0\right)$ and $\left(0 , 4\right]$. There is a local maximum at $x = - 4$ and a local minimum at $x = 4$.

#### Explanation:

Let $f \left(x\right) = \frac{16}{x} + x - 5 = 16 {x}^{- 1} + x - 5$. Then the derivative is $f ' \left(x\right) = - 16 {x}^{- 2} + 1 = 1 - \frac{16}{{x}^{2}}$. Setting this equal to zero gives $\frac{16}{{x}^{2}} = 1$ so that ${x}^{2} = 16$ and $x = \pm 4$. These are the two critical points.

The second derivative is $f ' ' \left(x\right) = 32 {x}^{- 3} = \frac{32}{{x}^{3}}$ so that $f ' ' \left(- 4\right) < 0$ and $f ' ' \left(4\right) > 0$. This means there is a local maximum value at $x = - 4$ and a local minimum value at $x = 4$ (by the Second Derivative Test).

This also could have been derived from the First Derivative Test, in which we can also see that $f ' \left(x\right) \setminus \ge q 0$ when $x \setminus \le q - 4$ and when $x \setminus \ge q 4$ and $f ' \left(x\right) \setminus \le q 0$ when $- 4 \setminus \le q x < 0$ and when $0 < x \setminus \le q 4$. This information also tells us that the function is increasing on the intervals $\left(- \infty , - 4\right]$ and $\left[4 , \infty\right)$ and is decreasing on the intervals $\left[- 4 , 0\right)$ and $\left(0 , 4\right]$.

The local maximum value at $x = - 4$ is $f \left(- 4\right) = - 4 - 4 - 5 = - 13$ and the local minimum value at $x = 4$ is $f \left(4\right) = 4 + 4 - 5 = 3$.

The graph of $f$ is shown below.

graph{16/x+x-5 [-80, 80, -40, 40]}