Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 2 - Chapter Test - Page 235: 17

Answer

(a) $2x^2-4x-2$ (b) $4x^2+4x-12$ (c) $-3x^4-12x^3+22x^2+28x-35$ (d) $\frac{3x^2-7}{-x^2-4x+5}$ (e) $3x^4+24x^2+18x^2-120x+68$ (f) $-9x^4+30x^2-16$

Work Step by Step

(a) $(f+g)(x)$ $=f(x)+g(x)$ $=3x^2-7+(-x^2-4x+5)$ $=3x^2-7-x^2-4x+5$ $=2x^2-4x-2$ (b) $(f-g)(x)$ $=f(x)-g(x)$ $=3x^2-7-(-x^2-4x+5)$ $=3x^2-7+x^2+4x-5$ $=4x^2+4x-12$ (c) $(fg)(x)$ $=f(x)g(x)$ $=(3x^2-7)(-x^2-4x+5)$ $=-3x^4-12x^3+22x^2+28x-35$ (d) $(f/g)(x)$ $=\frac{f(x)}{g(x)}$ $=\frac{3x^2-7}{-x^2-4x+5}$ (e) $(f\circ g)(x)$ $=f(g(x))$ $=3(-x^2-4x+5)^2-7$ $=3(x^4+8x^3+6x^2-40x+25)-7$ $=3x^4+24x^3+18x^2-120x+75-7$ $=3x^4+24x^2+18x^2-120x+68$ (f) $(g\circ f)(x)$ $=g(f(x))$ $=-(3x^2-7)^2-4(3x^2-7)+5$ $=-(9x^4-42x+49)-12x^2+28+5$ $=-9x^4+42x^2-49-12x^2+33$ $=-9x^4+30x^2-16$
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