What are the points of inflection, if any, of #f(x)=e^(2x) - e^x

#?

Question

What are the points of inflection, if any, of #f(x)=e^(2x) - e^x

#?

Emilia Aguilar · Accepted Answer

Crap.

Was complete garbage, so disregard what I said.

Evelyn Agard · Answer

There is an inflection point at #x=-2ln(2)#

We utilize the second derivative test in order to identify inflection points.

#f(x) = e^(2x) - e^(x)# #f'(x) = 2e^(2x) - e^(x)# #f''(x) = 4e^(2x) - e^(x)#

We apply the second derivative test by setting #f''(x)# equal to #0#.

#4e^(2x) - e^x = 0# #4e^(2x) = e^(x)# #ln(4e^(2x)) = ln(e^x)#

One characteristic of logarithms is their ability to convert terms that are multiplied in a single logarithm into a sum of logarithms for each term:

#ln(4e^(2x)) = ln(e^x)# #ln(4) + ln(e^(2x)) = ln(e^(x))# #ln(4) + 2x = x# #x = -ln(4)# #x=-ln(2^2)# #x=-2ln(2) ~~ -1.3863...#

Even though exponentials don't typically cause inflection points, the fact that one is being subtracted from the other implies that they might "affect" the graph in a way that makes an inflection point possible.

graph{-4.278, 1.88, -1.63, 1.447]} e^(2x) - e^(x)

graph: #f(x) = e^(2x) - e^(x)#

As you can see, the line appears to be concave down in the part to the left of the point, while it changes and becomes concave up in the portion to the right.

Axel Alvarez · Answer

undefined To find the points of inflection for $ f(x) = e^{2x} - e^x $:

1. Find the second derivative, $ f''(x) $.
2. Solve for $ x $ when $ f''(x) = 0 $ to find potential points of inflection.
3. Test the concavity of the function around these potential points using the second derivative test:
   - If $ f''(x) > 0 $ on an interval, the function is concave up.
   - If $ f''(x) < 0 $ on an interval, the function is concave down.
4. Confirm the concavity change to identify points of inflection.

After these steps, you will have determined the points of inflection, if any, for $ f(x) = e^{2x} - e^x $.