Consider the following function. (If an answer does not...

Consider the following function. (If an answer does not exist, enter DNE.) f(x) = 1 + 5/x - 7/x^2 (a) Find the vertical asymptote(s). (Enter your answers as a comma-separated list.) x = 0 Find the horizontal asymptote(s). (Enter your answers as a comma-separated list.) y = 1 (b) Find the interval where the function is increasing. (Enter your answer using interval notation.) (0, 7/15) Find the interval where the function is decreasing. (Enter your answer using interval notation.) (7/15, infinity) (c) Find the local maximum and minimum values. local maximum value local minimum value (d) Find the interval where the function is concave up. (Enter your answer using interval notation.) Find the interval where the function is concave down. (Enter your answer using interval notation.) Find the inflection point. (x, y) =

Transcript

00:01 So for this problem, we're given, it appears to be a rational function.

00:05 So even though the problem itself says 1 plus 5, it actually is clear that we are dealing with the function f of x equals 1 plus 5 over x minus 7 over x squared.

00:22 So we want to find vertical and horizontal asymptotes.

00:25 We also want to find where the intervals of increase and decrease.

00:28 And then also some maximum and minimum values, concave up, give down all of these kind of general parameters for analysis using calculus.

00:39 So what we're going to have here is first we want to find the vertical asymptote.

00:51 So we see that if we were to convert this into one complete function, we end up getting x squared plus 5x minus 7.

01:06 All over x squared.

01:08 And then because this x squared is in the denominator, to find the vertical asymptotes, we let the denominator equal zero and we get x equals zero.

01:18 So x equals zero is going to be our vertical asymptote.

01:23 And then for the horizontal asymptote, it's a little more complicated, but not too bad.

01:29 What we can do is take the limit as x goes to infinity of f of x.

01:36 So we see in this case that as x goes to infinity in the numerator and denominator, these two terms are going to dominate, but they're both x squared terms.

01:45 So we see that they're going to be approaching the same value.

01:48 So that's going to be 1.

01:49 So that means y equals 1 is going to be the horizontal asymptote.

01:55 Then we want to take the derivative, so f prime of x, and we want to set it equal to 0.

02:07 So this is going to give us a negative 5 over.

02:09 Over x squared plus 14 over x cubed, setting that equal to zero.

02:20 We end up getting, since our f of x is originally 1 plus 5 over x minus 7x squared, f prime of x is going to equal 0 right at negative 0 .709.

02:42 But before that it's going to be decrease or increasing.

02:50 So the interval we'll see is negative infinity to negative 0 .709 is an interval of increase.

03:02 These are going to be parentheses.

03:06 And then negative 0 .709 to infinity is going to be decrease.

03:18 And then there's also going to be a portion obviously at zero where it's doing neither.

03:25 So this is not including zero...

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