Finding Inflection Points (Applied Calc Question)

[Sam]

The problem: Find the inflection points, if any, for the following: f(x) = e^x + x^-1

I know to find inflection points I have to:

1. Compute f''(x)
2. Determine the points in the domain of f for which f''(x) = 0 or f''(x)
does not exist
3. Determine the sign of f''(x) to the left and right of each point x = c
found in step 2. If there is a change in the sign of f''(x) as we move
across the point x = c, then (c, f(c)) is an inflection point of f.

Well, this is what I came up with:

f'(x) = e^x -x^-2
f''(x)= e^x + 2x^-3

Then, I don't know what to do from there because e^x can never be zero, right? but I don't know. My teacher is saying there are inflection points...

Your help is much appreciated!

Sam

 

[HallsofIvy]

Yes, e^x is never 0, but an inflection point is NOT where "e^x= 0". It is where f"= e^x+ 2/x³= 0.

There is no "algebraic" way to solve that equation but it certainly has solutions: using Newton's method or a hand-dandy graphing calculator, we have a zero of f", and an inflection point, for x approximately 0.926.

 

[M98Ranger]

You are correct, e^x can never be equal to zero. However, we can still find inflection points by looking at the sign of f''(x) and whether it changes as we move across certain points in the domain.

First, let's find the points where f''(x) = 0 or does not exist. We can do this by setting f''(x) = 0 and solving for x.

f''(x) = e^x + 2x^-3 = 0
e^x = -2x^-3
Taking the natural logarithm of both sides:
x = ln(-2x^-3)

This equation has no real solutions, so there are no points in the domain where f''(x) = 0.

Next, let's look at the sign of f''(x) to the left and right of certain points in the domain. We can do this by plugging in values for x that are slightly greater and slightly smaller than the points we found in step 2.

For example, let's look at x = 0.1 and x = -0.1.

f''(0.1) = e^0.1 + 2(0.1)^-3 = 1.105
f''(-0.1) = e^-0.1 + 2(-0.1)^-3 = 0.895

Since f''(0.1) is positive and f''(-0.1) is negative, there is a change in sign as we move across x = 0. This means that (0, f(0)) is an inflection point of f(x).

We can repeat this process for other points in the domain, such as x = 1 and x = -1. Doing so, we can find that (1, f(1)) and (-1, f(-1)) are also inflection points of f(x).

Therefore, the inflection points for f(x) = e^x + x^-1 are (0, f(0)), (1, f(1)), and (-1, f(-1)).

I hope this helps! Keep in mind that finding inflection points can be tricky and may require some trial and error, so it's always a good idea to double check your work and ask for help if needed.