If c (x) = StartFraction 5 Over x minus 2 EndFraction and d(x) = x + 3, what is the domain of (cd)(x)?
2 answers:
Answer:
The domain of the function (cd)(x) will be all real values of x except x = 2.
Step-by-step explanation:
The two functions are and d(x) = x + 3
So, (cd)(x) =
Then, for x = 2 the function (cd)(x) will be undefined as zero in the denominator will make the function (cd)(x) undefined.
Therefore, the domain of the function (cd)(x) will be all real values of x except x = 2. (Answer)
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So hmmm check the picture below
so... the vertex is "p" distance from the focus and the directrix, thus, the vertex is really half-way between both
in this case, 2 units up from the focus or 2 units down from the directrix, and thus it lands at 3,3
now, the "p" distance is 2, however, the directrix is up, the focus point is below it, the parabola opens towards the focus point, thus, the parabola is opening downwards, and the squared variable is the "x"
because the parabola opens downwards, "p" is negative, and thus, -2
now, let's plug all those fellows in then
so... the vertex is "p" distance from the focus and the directrix, thus, the vertex is really half-way between both
in this case, 2 units up from the focus or 2 units down from the directrix, and thus it lands at 3,3
now, the "p" distance is 2, however, the directrix is up, the focus point is below it, the parabola opens towards the focus point, thus, the parabola is opening downwards, and the squared variable is the "x"
because the parabola opens downwards, "p" is negative, and thus, -2
now, let's plug all those fellows in then