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 = Find the horizontal asymptote(s). (Enter your answers as a comma-separated list.) y = (b) Find the interval where the function is increasing. (Enter your answer using interval notation.) Find the interval where the function is decreasing. (Enter your answer using interval notation.) (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) = () (e) Use the information from parts (a)-(d) to sketch the graph of f.

Transcript

00:01 You have f of x is equal to 1 plus 5 over x minus 7 over x squared.

00:10 First you want to find the vertical asymptotes.

00:13 The vertical asymptotes are where the function does not exist.

00:17 So the limit as it approaches that value will go to positive or negative infinity.

00:23 That is that x is equal to 0.

00:25 If x cannot be equal to 0 or you have a division by 0 error.

00:33 For horizontal asymptotes, horizontal asymptotes, we would find the limit as x approaches either positive or negative infinity.

00:42 I'm going to use positive infinity of 1 plus 5 over x minus 7 over x squared.

00:52 So as x goes to infinity, this goes to zero, this also goes to zero.

00:57 So this is equal to 1.

00:57 So x equals 1 is a horizontal asymptote.

01:07 That was still part a.

01:10 So then part b.

01:14 Find the interval where the function is increasing and where it's decreasing.

01:22 First, it's good to go ahead and write out the domain.

01:25 So remember, x cannot be equal to zero.

01:27 So we will need to include that when we're looking at intervals of increasing and decreasing and also concavity.

01:40 So first, take the first derivative.

01:48 So f prime of x is equal to the derivative of 1 is 0.

01:56 Then this would be negative 5 over x squared plus 14 over x cubed.

02:06 Set it equal to 0 to find the critical points.

02:19 So we can rearrange this to have a common denominator.

02:24 This would be 14 minus 5x over x cubed.

02:30 If we're going to set this equal to 0, that means x is 14 over 5 as a critical point.

02:41 And then to find where it's increasing or decreasing, we need to deal with zero, and we also need to deal with 14 -fifths.

02:51 So anything less than zero, the derivative is negative.

02:58 Between 0 and 14 -fifths, the derivative is positive.

03:02 Outside of 14 -fifths, the derivative is negative.

03:07 So it is decreasing from negative -infinity.

03:13 To zero, and then from 14 fifths to infinity, and it is increasing from zero to 14 fifths.

03:30 Find the local maximum and minimum...

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