Question

What is the inflection point of the equation `(x^5)(ln(x))`?

Answer 1

Let `f(x)=x^{5}ln(x)`, then the Product Rule says that `f'(x)=5x^{4}ln(x)+x^{5}\cdot \frac{1}{x}=5x^{4}ln(x)+x^{4}`. Now use the Product Rule again to get `f''(x)=20x^{3}ln(x)+5x^{4}\cdot \frac{1}{x}+4x^{3}=20x^{3}ln(x)+9x^{3}=x^{3}(20ln(x)+9)`.

These equations are only valid for `x>0` since `ln(x)` is undefined for `x\leq 0`. Therefore, `f''(x)=0` when `20ln(x)+9=0`, which means `\ln(x)=-\frac{9}{20}` and `x=e^{-9/20}\approx 0.6376`. In fact, you can check that `f''(x)` changes sign (from negative to positive) as `x` increases through the value `e^{-9/20}`, making `x=e^{-9/20}` the `x`-coordinate of a true inflection point of `f` (the graph of `f` will change from concave down to concave up as `x` increases through this value).