Asked by Steve
use f(x)=square root of 5-x and g(x)=x^2 + 2x
a-find f(g(x)), state the domain
b-find g(f(x)),state the domain
c-find g(f(-1))
d-find f(x)/g(x), state the domain, then evaluate for x=3
a-find f(g(x)), state the domain
b-find g(f(x)),state the domain
c-find g(f(-1))
d-find f(x)/g(x), state the domain, then evaluate for x=3
Answers
Answered by
Reiny
f(x) = √(5-x) , g(x) = x^2 + 2x
f(g(x)
= √(5 - (x^2+2x) )
= √(5 - x^2 - 2x)
for this to be real, 5 - x^2 - 2x ≥ 0
x^2 + 2x - 5 ≤ 0
solve x^2 + 2x - 5 = 0 ,
the domain will be all values of x <b>between</b> the two solution values
g(f(x)
= (√(5-x))^2 + 2√(5-x)
= 5 - x + 2√(5-x)
domain is :
5-x ≥ 0
-x ≥ -5
x ≤ 5
you try c and d
f(g(x)
= √(5 - (x^2+2x) )
= √(5 - x^2 - 2x)
for this to be real, 5 - x^2 - 2x ≥ 0
x^2 + 2x - 5 ≤ 0
solve x^2 + 2x - 5 = 0 ,
the domain will be all values of x <b>between</b> the two solution values
g(f(x)
= (√(5-x))^2 + 2√(5-x)
= 5 - x + 2√(5-x)
domain is :
5-x ≥ 0
-x ≥ -5
x ≤ 5
you try c and d
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