Suppose f(x)=|P(x)|. Determine the range of f
(1,-3), (4,1), (3,3), (2,4), (-2,4), (-3,5), (-1,6)
Algebra 1 - Damon today at 10:12am
Nothing can be negative.
No y can be different from x
Can you explain further please
If f(x) is equal to the absolute value of another quanity, then no f(x) can be negative, by definition. (1,-3) f(x) cannot be -3.
If the points given are those of P(x), then f(x) has the same values, with the exception of (1,-3). f(x) contains (1,3) since |-3|=3, and the range of f(x) is all the distinct values of y.
Sure! To determine the range of the function f(x) = |P(x)|, we first need to understand what the absolute value function does. The absolute value of a number is the distance of that number from zero on the number line. Since the absolute value of any number is never negative, the function |P(x)| will always be non-negative or zero.
Now let's consider the given points in the form (x, y). We need to find the corresponding range values of the function f(x) = |P(x)| for each of these points.
(1, -3): Here, P(1) = -3, and the absolute value of -3 is 3. So f(1) = |P(1)| = 3.
(4, 1): Here, P(4) = 1, so f(4) = |P(4)| = 1.
(3, 3): Here, P(3) = 3, so f(3) = |P(3)| = 3.
(2, 4): Here, P(2) = 4, so f(2) = |P(2)| = 4.
(-2, 4): Here, P(-2) = 4, so f(-2) = |P(-2)| = 4.
(-3, 5): Here, P(-3) = 5, so f(-3) = |P(-3)| = 5.
(-1, 6): Here, P(-1) = 6, so f(-1) = |P(-1)| = 6.
From these calculations, we can see that the range of f(x) = |P(x)| is {0, 1, 3, 4, 5, 6}. These are the non-negative or zero values associated with the function for the given points.