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Ch. 2 - Functions and Graphs
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 38

Find f+gf+g, f−gf-g, fgfg, and fg\(\frac{f}{g}\). Determine the domain for each function.
f(x)=5−x2f\(\left\)(x\(\right\))=5-x^2, g(x)=x2+4x−12g\(\left\)(x\(\right\))=x^2+4x-12

Guida verificata passo dopo passo
1
Step 1: To find \( (f+g)(x) \), add the functions \( f(x) \) and \( g(x) \). This means you will add \( 5 - x^2 \) and \( x^2 + 4x - 12 \). Combine like terms to simplify.
Step 2: To find \( (f-g)(x) \), subtract \( g(x) \) from \( f(x) \). This involves subtracting \( x^2 + 4x - 12 \) from \( 5 - x^2 \). Again, combine like terms to simplify.
Step 3: To find \( (fg)(x) \), multiply the functions \( f(x) \) and \( g(x) \). This requires distributing \( 5 - x^2 \) across \( x^2 + 4x - 12 \) and combining like terms.
Step 4: To find \( \left(\frac{f}{g}\right)(x) \), divide \( f(x) \) by \( g(x) \). This means writing \( \frac{5 - x^2}{x^2 + 4x - 12} \). Simplify if possible, and identify any restrictions on the domain where the denominator is zero.
Step 5: Determine the domain for each function. For \( f+g \) and \( f-g \), the domain is all real numbers. For \( fg \), the domain is also all real numbers. For \( \frac{f}{g} \), exclude values that make the denominator zero by solving \( x^2 + 4x - 12 = 0 \) to find the restricted values.

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