Question

What are the points of inflection of `f(x)= e^(2x) - e^x`?

Answer 1

`x=-ln4`
`xapprox -1.3863`

Explanation:

Points of inflection occur when the second derivative is equal to zero. Find this by first differentiating `f(x)` to get `f'(x)`, then differentiating `f'(x)` to get `f''(x)`.

`f(x)=e^(2x)-e^x`

`f'(x)=2e^(2x)-e^x`

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

Set `f''(x)` equal to zero to find possible points of inflection:
`0=4e^(2x)-e^x`
`e^x=4e^(2x)`
Rewrite as a natural log:
`x=ln(4e^(2x))`
`x=ln4 + ln(e^(2x))`
`x=ln4+2xlne`
`x=ln4+2x`
`-x=ln4`

`x=-ln4`
`xapprox -1.3863`
Check if this is a point of inflection by making sure `f''(x)` is positive on one side of the x value, and negative on the other (make a sign chart):
`"- "x=-ln4" +"`

Therefore, `x=-ln4` is the point of inflection of `f(x)=e^(2x)-e^x`