30 POINTS 7TH GRADE MATH! There are 15 girls in your class this year. That is a 15% increase from last year.
1 answer:
You might be interested in
Answer:
See below
Step-by-step explanation:
I assume the function is
A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, is the only vertical asymptote.
B) Set the first derivative equal to 0 and solve:
Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:
Therefore, the function increases on the interval and decreases on the interval
.
C) Since we determined that the slope is 0 when from the first derivative, plugging it into the original function tells us where the extrema are. Therefore,
, meaning there's an extreme at the point
, but is it a maximum or minimum? To answer that, we will plug in
into the second derivative which is
. If
, then it's a minimum. If
, then it's a maximum. If
, the test fails. So,
, which means
is a local maximum.
D) Now set the second derivative equal to 0 and solve:
We then test where is negative or positive by plugging in test values. I will use -1 and 3 to test this:
, so the function is concave down on the interval
, so the function is concave up on the interval
The inflection point is where concavity changes, which can be determined by plugging in into the original function, which would be
, or
.
E) See attached graph
