Answer:
{ x | 1, 2, 3, 4, 5, 6, 7, 8, 9 }
Step-by-step explanation:
We note that points between integer values have been excluded, so domain consists of integer x-values from 1 to 9, or
{ x | 1, 2, 3, 4, 5, 6, 7, 8, 9 }
and <em>not</em>
1 <= x <= 9
Answer:
18 apples equals 6 bananas
Answer:
10
Step-by-step explanation:
1x^3-2x^2-4x-6
x^3-2x^2-4x-6
(4)^3-2(4)^2-4(4)-6
64-2(16)-16-6
64-32-16-6
32-16-6
16-6
10
So when x=4, the output will be 10
E) 2
Remember that the first derivative of a function is the slope of the function at any specified point. We've been told that f(0) = -5 and that f'(x) is always less than or equal to 3. So let's look at the available options and see what the average slope would have to be in order to get the specified value of f(2).
A) -10: (-10 - -5)/(2 - 0) = -5/2 = -2.5
B) -5: (-5 - -5)/(2 - 0) = 0/2 = 0
C) 0: (0 - -5)/(2 - 0) = 5/2 = 2.5
D) 1: (1 - -5)/(2 - 0) = 6/2 = 3
E) 2: (2 - -5)/(2 - 0) = 7/2 = 3.5
Now taking into consideration the mean value theorem, the value of the function f'(x) has to have the value equal to the average slope between the two points at at least one point between the two given values. For options A, B, C, and D it's possible for f'(x) to return values that make that slope possible. However, for option E, the mean value theorem indicates that f'(x) has to have the value of 3.5 for at least 1 point between x=0 and x=2. And since we've been told that f'(x) is less than or equal to 3 for all possible values of x, that is in conflict and f(2) can not have the value of 2.
