## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 10 - Exponents and Radicals - Mid-Chapter Review - Mixed Review: 7

#### Answer

$f(-5)=-4$

#### Work Step by Step

Replacing $x$ with $-5$ in $f(x)=\sqrt[3]{12x-4},$ then \begin{array}{l}\require{cancel} f(-5)=\sqrt[3]{12(-5)-4} \\\\ f(-5)=\sqrt[3]{-60-4} \\\\ f(-5)=\sqrt[3]{-64} \\\\ f(-5)=\sqrt[3]{(-4)^3} \\\\ f(-5)=-4 .\end{array}

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