Asked by Anonymous
Find f of g of h and state the exact domain. f(x) = sqrt x
g(x)= e^x - 13e^x/2 + 42
h(x) = 2x
I need help on this one bad! Please help!
g(x)= e^x - 13e^x/2 + 42
h(x) = 2x
I need help on this one bad! Please help!
Answers
Answered by
Steve
domain of f is all x≥0
domain of g = all reals
domain of h = all reals
f(g(h)) = f(g(2x))
= f(e^2x - 13e^x + 42)
= sqrt(e^2x - 13e^x + 42)
So, where is e^2x - 13e^x + 42 ≥ 0?
(e^x - 7)(e^x - 6) ≥ 0
So, either e^x - 7 ≥ 0 and e^x - 6 ≥ 0
or e^x - 7 ≤ 0 and e^x - 6 ≤ 0
e^x ≥ 7 and e^x ≥ 6 means e^x ≥ 7, or x ≥ ln 7
e^x ≤ 7 and e^x ≤ 6 means e^x ≤ 6, or x ≤ ln 6
SO, f(g(h(x))) ≥ 0 everywhere except (ln6,ln7)
domain of g = all reals
domain of h = all reals
f(g(h)) = f(g(2x))
= f(e^2x - 13e^x + 42)
= sqrt(e^2x - 13e^x + 42)
So, where is e^2x - 13e^x + 42 ≥ 0?
(e^x - 7)(e^x - 6) ≥ 0
So, either e^x - 7 ≥ 0 and e^x - 6 ≥ 0
or e^x - 7 ≤ 0 and e^x - 6 ≤ 0
e^x ≥ 7 and e^x ≥ 6 means e^x ≥ 7, or x ≥ ln 7
e^x ≤ 7 and e^x ≤ 6 means e^x ≤ 6, or x ≤ ln 6
SO, f(g(h(x))) ≥ 0 everywhere except (ln6,ln7)
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