Question 1054882: Given the function y = f(x) = -x2 + 8x – 12. Suppose the domain of the function is restricted to the semi-open interval (3, 7]. What is the corresponding range of the function in interval notation?
a. (-5, 4)
b. [-5, 4]
c. [-5, 3]
d. [-5, 3)
e. [-4, 5]
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the equation is y = -x^2 + 8x - 12.
the domain is (3,7]
this means that x is > 3 and x <= 7
this can be written as 3 < x <= 7
in interval notation, that looks like (3,7].
the value of y is defined when x > 3 and when x <= 7.
the function is y = -x^2 + 8x - 12.
this is a quadratic function and has a maximum or minimum value at x = -b/2a.
since a = -1 and b = 8 and c = -12, then:
x = -b/2a = -8/-2 = 4.
the function has a maximum value at x = 4.
when x = 7, the value of the function is y = -5.
when x = 4, the value of the function is y = 4
the range of y goes from the max of y to the min of y, within the confines of the domain.
this means that y goes from 4 to -5.
put the minimum value ahead of the maximum value and this means that y goesx from -5 to 4.
x = 4 and x = 7 are both valid values within the domain.
therefore y = 4 and y = -5 are both valid values within the range.
this means that y is >= -5 and y is <= 4.
this can be written as -5 <= x <= 4.
in interval notation, this can be written as [-5,4]
that would be selection b.
if you graph the equation, you will see that this makes sense.
here's the graph.
even though it's shown on the graph, the coordinate point of (3,3) is not a valid point of the equation because x = 3 is not part of the domain.
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