Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.1 - Functions - Exercises: 28

Answer

Evaluate $k(x) = 2x^3 - 3x^2$: $k(0) = 0$ $k(3) = 27$ $k(-3) =-81$ $k(\frac{1}{2}) = -\frac{1}{2}$ $k(\frac{a}{2}) =\frac{a^3-3a^2}{4}$ $k(-x) = -2x^3 - 3x^2$ $k(x^3) =2x^9 - 3x^6$

Work Step by Step

For x = 0 $k(0) = 2(0)^3 - 3(0)^2$ ... evaluate exponents $ = 2(0) - 3(0)$ ... multiply $=0 - 0 = 0$ ____________________ For x = 3 $k(3) = 2(3)^3 - 3(3)^2$ ... evaluate exponents $ = 2(27) - 3(9)$ ... multiply $=54 - 27 = 27$ ____________________ For x = -3 $k(-3) = 2(-3)^3 - 3(-3)^2$ ... evaluate exponents $ = 2(-27) - 3(9)$ ... multiply $=-54 - 27 = -81$ ____________________ For x = $\frac{1}{2}$ $k(\frac{1}{2}) = 2(\frac{1}{2})^3 - 3(\frac{1}{2})^2$ ... distribute exponents $ = 2(\frac{1^3}{2^3}) - 3(\frac{1^2}{2^2})$ ... evaluate exponents $=2(\frac{1}{8}) - 3(\frac{1}{4})$ ... multiply $=\frac{2}{8} - \frac{3}{4}$ ... simplify for common denominator $=\frac {1}{4} - \frac{3}{4}$ ... subtract $=-\frac{2}{4}= - \frac{1}{2}$ ____________________ For x = $\frac{a}{2}$ $k(\frac{a}{2}) = 2(\frac{a}{2})^3 - 3(\frac{a}{2})^2$ ... distribute exponents $ = 2(\frac{a^3}{2^3}) - 3(\frac{a^2}{2^2})$ ... evaluate exponents $=2(\frac{a^3}{8}) - 3(\frac{a^2}{4})$ ... multiply $=\frac{2a^3}{8} - \frac{3a^2}{4}$ ... simplify for common denominator $=\frac {a^3}{4} - \frac{3a^2}{4}$ ... subtract $=\frac{a^3-3a^2}{4}$ ____________________ For x = $-x$ $k(-x) = 2(-x)^3 - 3(-x)^2$ ... evaluate exponents $ = 2(-x^3) - 3(x^2)$ ... multiply $=-2x^3 - 3x^2$ ____________________ For x = $x^3$ $k(x^3) = 2(x^3)^3 - 3(x^3)^2$ ... evaluate exponents $ = 2(x^9) - 3(x^6)$ ... multiply $=2x^9 - 3x^6$ ____________________
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