Blog: `y = x ln(x)` Locate any relative extrema and points of inflection.

We are asked to locate any relative extrema or inflection points for the graph of $\displaystyle{y}={x}{\ln{{x}}}$ :

The domain of the function is x>0.

Extrema can only occur at critical points; that is when the first derivative is zero or fails to exist.

$\displaystyle{y}'={\ln{{x}}}+{x}\cdot\frac{{1}}{{x}}=\Rightarrow{y}'={\ln{{x}}}+{1}$

This function is continuous and differentiable for all x in the domain, so setting y'=0 we get:

$\displaystyle{\ln{{x}}}+{1}={0}=\Rightarrow{\ln{{x}}}=-{1}=\Rightarrow{x}=\frac{{1}}{{e}}\approx{0.368}$

For 0<x<1/e...

We are asked to locate any relative extrema or inflection points for the graph of $\displaystyle{y}={x}{\ln{{x}}}$ :

The domain of the function is x>0.

Extrema can only occur at critical points; that is when the first derivative is zero or fails to exist.

$\displaystyle{y}'={\ln{{x}}}+{x}\cdot\frac{{1}}{{x}}=\Rightarrow{y}'={\ln{{x}}}+{1}$

This function is continuous and differentiable for all x in the domain, so setting y'=0 we get:

$\displaystyle{\ln{{x}}}+{1}={0}=\Rightarrow{\ln{{x}}}=-{1}=\Rightarrow{x}=\frac{{1}}{{e}}\approx{0.368}$

For 0<x<1/e the first derivative is negative and for x>1/e it is positive, so the only extrema is a minimum at x=1/e.

Inflection points can only occur when the second derivative is zero:

$\displaystyle{y}''=\frac{{1}}{{x}}>{0}\forall{x}$ so there are no inflection points.

The graph:

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