Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Cumulative Review - Page 373: 2

Answer

$(a) \quad f(3) = 10 $ $(b) \quad f(-x)= 2x^{2}+3x+1 $ $(c)\quad f(x+h) =2x^{2}+(4h-3)x+2h^{2}-3h+1$

Work Step by Step

Given: $f(x)=2 x^{2}-3 x+1$ (a) To find $f(3)$, substitute $x=3$ to obtain: \begin{align*} f(3)&=2(3)^{2}-3(3)+1\\ f(3)&=18-9+1\\ f(3)&=10 \end{align*} (b) To find $f(-x)$, replace $x$ by $-x$: \begin{align*}f(-x) &=2(-x)^{2}-3(-x)+1 \\ f(-x)&=2 x^{2}+3 x+1 \end{align*} (c) To find $f(x+h)$, replace $x$ by $(x+h)$ to obtain \begin{align*} f(x+h)&= 2(x+h)^{2}-3(x+h)+1 \\ f(x+h)&= 2\left(x^{2}+2 x h+h^{2}\right) -3 x-3 h+1 \\ f(x+h)&=2 x^{2}+4 x h+2 h^{2} -3 x-3 h+1 \\ f(x+h)&=2 x^{2}+(4h-3)x +2 h^{2}-3 h+1 \end{align*}
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