tim has a large collection of books. the number of paperback books he owns is three less than four times the number of hardcover
2 answers:
We can solve this by setting up a couple of equations.
Let's allow x to represent the number of paperbacks Tim owns, and allow y to represent the number of hardcover books he owns.
Using the information in the question, we can write the equations:
1)
2)
Let's rearrange equation 1 so that it is in standard form:
And then let's multiply equation 2 by 4 so that we can cancel out y when we solve the system of equations:
Then we can add the two equations and solve for x:
1)
+ 2)
------------------------------------
So now we now the number of paperback books Tim has is 357. Let's plug this into one of the original equations to solve for the number of hardcover books (y):
And now we know that Tim owns 90 hardcover books.
Let's allow x to represent the number of paperbacks Tim owns, and allow y to represent the number of hardcover books he owns.
Using the information in the question, we can write the equations:
1)
2)
Let's rearrange equation 1 so that it is in standard form:
And then let's multiply equation 2 by 4 so that we can cancel out y when we solve the system of equations:
Then we can add the two equations and solve for x:
1)
+ 2)
------------------------------------
So now we now the number of paperback books Tim has is 357. Let's plug this into one of the original equations to solve for the number of hardcover books (y):
And now we know that Tim owns 90 hardcover books.
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A) 
is continuous, so 
is also continuous. This means if we were to integrate 
, the same constant of integration would apply across its entire domain. Over 
, we have 
. This means that
For to be continuous, we need the limit as 
to match 
. This means we must have
Now, over
, we have 
, so 
, which means 
.
b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,
c) is increasing when 
, and concave upward when 
, i.e. when is also increasing.
We have
over the intervals 
and 
. We can additionally see that 
only on and .
d) Inflection points occur when
, and at such a point, to either side the sign of the second derivative 
changes. We see this happening at 
, for which 
, and to the left of we have decreasing, then increasing along the other side.
We also have along the interval 
, but even if we were to allow an entire interval as a "site of inflection", we can see that to either side, so concavity would not change.
For
Now, over
b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,
c)
We have
d) Inflection points occur when
We also have