If you would like to solve for f(g(x)) when x = 1, you can do this using the following steps:
<span>f(x) = x^2 - 3x + 6
g(x) = x - 3/2
f(g(x)) = f(</span>x - 3/2) = (x - 3/2)^2 - 3 * (x - 3/2) + 6
x = 1
f(g(1)) = f(1 - 3/2) = f(-1/2) = (-1/2)^2 - 3 * (-1/2) + 6 = 1/4 + 3/2 + 6 = 1/4 + 6/4 + 24/4 = 31/4 = 7 3/4
The correct result would be 7 3/4. add me as a friend
if you look at the part where the first part connects with the second part:
y = 5 if x < - 2
y = -2x + 1 if -2 ≤ x < 1
we don't have a discontinuity there, so there shouldn't be a dot.
<h3>
</h3><h3>
What is wrong with the graph?</h3>
When we graph over intervals like (a, b) or [a, b] or something like that, we use dots to define the end of the intervals, and to denote that the function ends abruptly or we have a jump.
In this case, you can see that between the end and the second part and the beginning of the third part there is a jump, so the use of dots is correct there, but if you look at the part where the first part connects with the second part:
y = 5 if x < - 2
y = -2x + 1 if -2 ≤ x < 1
we don't have a discontinuity there, so there shouldn't be a dot.
That is the only error with the graph.
If you want to learn more about piecewise functions:
brainly.com/question/3628123
#SPJ1
Answer:
Rhombus
Step-by-step explanation:
Answer:
-1/2
Step-by-step explanation:
y2-y1 4-5 -1
--------- = -------- = --------
x2-x1 6-4 2
