Math 1B Homework Problem 5.4#18
This problem belongs to a larger class of problems: Find the interval on which the function is concave upward. Assume that a and k are both positive.
Solution:
An example of this is 5.4#18: . While it’s not essential to the problem, it’s nice to find an antiderivative using trig substitution. Note the sum of squares (well, 3 is the square of its square root) so we’d like to have , which we can obtain by substituting , or, more simply, whence and the integral becomes
This is just a transformation of the arctan function that, among other things, shifts its inflection point from the origin to the point . On the TI-89 (or Voyage 200) we can plot the function by first defining it like so:
The plot comes out all right, but only after an eternity of waiting:
Alternatively, you can do the integral:
which gives the same plot, only much faster:
Note that the integral function goes through the origin. Do you see why that makes sense?