Question

How do you find the inflection point of the function `f(x) = x9ln(x)`?

Answer 1

Assuming that this is `f(x) = x^9lnx`, see the explanation section, below

Explanation:

`f(x) = x^9lnx`,

Note that: `"dom"(f) = (0,oo)`

`f'(x) = 9x^8lnx+x^8`

`f''(x) - 72x^7lnx + 9x^7 + 8x^7 = x^7(72lnx+17)`

`f''(x)` is undefined at `x=0`, but there cannot be an inflection point at `x=0` `" "`(Because `"dom"(f) = (0,oo)`)

`f''(x) = 0` at `lnx = -17/72`, so `x = e^(-17/72)`

On the interval `(0,e^(-17/72))`,

`x^7` is positive and `(72lnx+17)` is negative.

(Recall that `lnx` is increasing, so
`x < e^(-17/72) rArr lnx < ln (e^(-17/72)) = -17/72`

`rArr 72lnx < -17`

`rArr 72lnx+17 < 0`)

So, on the interval `(0,e^(-17/72))`, we have `f''(x)` is negative.

By similar reasoning, on the interval `(e^(-17/72),oo)`, we have `f''(x)` is positive.

Therefore, the concavity changes at `x= e^(-17/72)`, which is in the domain of `f` and finally, we conclude that there is one inflection point.

The inflection point is `(e^(-17/72), f(e^(-17/72))) = (e^(-17/72), -17/72(e^(-17/8)))`

(Do the arithmetic if someone insists.)