For the functions f(x)=x−3 and g(x)=x+2, find and determine the domain in each case: (a)gf (b)f+g

f(x)=x−3,g(x)=x+2 a) gf=x+2x−3 =(x+2)(x−3) Since, square root is not defined for negative values ⇒(x+2)(x−3)≥0 ⇒x≥3 and x≤−2 ⇒ Domain =(−∞,−2]∪[3,∞) b) g+f=x+2+x−3 x+2 is defined for x+2≥0 x≥−2(1) x−3 is defined for x−3≥0 x≥3(2) Combining (1) and (2) for x remain positive ⇒x≥3 ⇒ Domain: =[3,∞)

Expert answers to "For the functions f(x)=x−3 and g(x)=x+2, find and..."

High School
Algebra
Pre-Algebra
Factors and multiples

Jerold

Answered question

2021-08-10

For the functions

f (x) = \sqrt{x - 3}

and

g (x) = \sqrt{x + 2}

, find and determine the domain in each case: (a)gf (b)

f + g

Answer & Explanation

Viktor Wiley

Skilled2021-08-11Added 84 answers

f (x) = \sqrt{x - 3}, g (x) = \sqrt{x + 2}

g f = \sqrt{x + 2} \sqrt{x - 3}

= \sqrt{(x + 2) (x - 3)}

Since, square root is not defined for negative values

\Rightarrow (x + 2) (x - 3) \geq 0

\Rightarrow x \geq 3

and

x \leq - 2

\Rightarrow

Domain

= (- \infty, - 2] \cup [3, \infty)

g + f = \sqrt{x + 2} + \sqrt{x - 3}

\sqrt{x + 2}

is defined for

x + 2 \geq 0

x \geq - 2 (1)

\sqrt{x - 3}

is defined for

x - 3 \geq 0

x \geq 3 (2)

Combining (1) and (2) for x remain positive

\Rightarrow x \geq 3

\Rightarrow

Domain:

= [3, \infty)

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