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Question 1168075: Let f(x)=√42−x and g(x)=x²−x.
Give the domain of f ∘ g using interval notation.
I can't figure this one yet, please!!
Found 2 solutions by solver91311, Theo:
Answer by solver91311(24713)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! f(x) = sqrt(42 - x)
g(x) = x^2 - x
fog(x) = f(g(x)) = f(x^2-x) = (42 - (x^2 - x))
you are replacing the x in f(x) with x^2 - x from g(x).
to get a real answer, the expression under the square root sign has to be greater than or equal to 0.
in order for that to happen, 42 - (x^2 - x) must be greater than or equal to 0.
your inequality becomes 42 - (x^2 - x) >= 0
simplify to get 42 - x^2 + x >= 0
order the terms in descending order of degree to get = -x^2 + x + 42 >= 0
solve this quadratic equation for -x^2 + x + 42 = 0
to factor, multiply both sides of the equation by -1 to get x^2 - x + 42 = 0
factor to get (x - 7) * (x + 6) = 0
solve for x to get x = 7 or x = -6
when x = 7, 42 - (x^2 - x) becomes 42 - (49 - 7) which becomes 42 - 42 which becomes 0.
when x = -6, 42 - (x^2 - x) becomes 42 - (36 + 6) which becomes 42 - 42 which becomes 0.
you get 42 - (x^2 - x) = 0 when x = 7 or when x = -6
you want it to be greater than 0 as well.
when x = 6 (less than 7), 42 - (x^2 - x) becomes 42 - (36 - 6) which becomes 42 - 30 which is greater than 0.
this means x <= 7 is good.
when x = -5 (greater than -6), 42 - (x^2 - x) becomes 42 - (25 + 5) which becomes 42 - 30 which is greater than 0.
this means x >= -6 is good.
it appears that you will get a real solution when -6 <= x <= 7.
that should be your domain.
your equation can be graphed as shown below.
the domain is -6 <= x <= 7
the range is 0 <= y <= 6.5
that's because sqrt(42 - sqrt(x^2 -x)) will be at a maximum when (x^2 - x) is at a minimum.
x^2 - x is a quadratic equation whose minimum will be when x = -b/2a
a is the coefficient of the x^2 term and b is the coefficient of the x term.
a = 1 and b = -1, so x = -b/2a becomes x = 1/2 = .5
x^2 - x will be at a minimum when x = .5
when x = .5, sqrt(42 - (x^2 - x)) becomes sqrt(42 - (.5^2 - .5)) which becomes sqrt(42 - (.25 - .5)) which becomes sqrt(42 - (-.25)) which becomes sqrt(42 + .25) which becomes sqrt(42.25) which is equal to 6.5.
the maximum y value is 6.5
the range is 0 <= y <= 6.5
you have:
the domain is -6 <= x <= 7 and the range is 0 <= y <= 6.5
in interval notation, the domain is x = [-6,7].
if fog(x) is a good equation, then you should get the same answer as when you solve for g(x) and then use the answer from g(x) as input to f(x)
for example:
when x = 5, g(x) = x^2 - x becomes g(x) = 25 - 5 which becomes g(x) = 20
g(x) becomes x in f(x).
when x = 20, f(x) = sqrt(42 - x) becomes f(x) = sqrt(42 - 20) which becomes f(x) = sqrt(22)
when x = 5, fog(x) = sqrt(42 - (x^2 - x)) becomes sqrt(42 - (25 - 5)) which becomes sqrt(42 - 20) which becomes sqrt(22).
you get the same value whether you use the fog(x) equation or whether you first find g(x) and then use the result from g(x) as input to f(x).
this confirms the solution is correct.
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