Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises - Page 278: 101

$16x^{2}$

According to the definition of negative exponents, $a^{-n}=\frac{1}{a^{n}}$ (where $a\ne0$). Therefore, $(\frac{1}{4x})^{-2}=\frac{1}{(\frac{1}{4x})^{2}}=(\frac{4x}{1})^{2}=\frac{4^{2}x^{2}}{1^{2}}=\frac{16x^{2}}{1}=16x^{2}$.