`y = x^2 ln(x/4)` Locate any relative extrema and points of inflection.

Sunday, September 9, 2012

$\displaystyle{y}={{x}}^{{2}}{\ln{{\left(\frac{{x}}{{4}}\right)}}}$ Locate any relative extrema and points of inflection.

Locate any extrema and points of inflection for the graph of $\displaystyle{y}={{x}}^{{2}}{\ln{{\left(\frac{{x}}{{4}}\right)}}}$ :

The domain for the function is x>0.

Extrema can only occur at critical points, or where the first derivative is zero or fails to exist.

$\displaystyle{y}'={2}{x}{\ln{{\left(\frac{{x}}{{4}}\right)}}}+{{x}}^{{2}}{\left(\frac{{\frac{{1}}{{4}}}}{{\frac{{x}}{{4}}}}\right)}$

$\displaystyle{y}'={2}{x}{\ln{{\left(\frac{{x}}{{4}}\right)}}}+{x}$ This is continuous and differentiable for all x in the domain so we set it equal to zero:

$\displaystyle{2}{x}{\ln{{\left(\frac{{x}}{{4}}\right)}}}+{x}={0}=\Rightarrow{\ln{{\left(\frac{{x}}{{4}}\right)}}}=-\frac{{1}}{{2}}$

$\displaystyle\frac{{x}}{{4}}={{e}}^{{-\frac{{1}}{{2}}}}=\Rightarrow{x}={4}{{e}}^{{-\frac{{1}}{{2}}}}\approx{2.43}$

For 0<x<4e^(-1/2) the first derivative is negative, greater it is positive so there is a minimum at $\displaystyle{x}={4}{{e}}^{{-\frac{{1}}{{2}}}}$ which is the only extrema.

Any inflection points can only occur if the second derivative is zero:

$\displaystyle{y}''={2}{\ln{{\left(\frac{{x}}{{4}}\right)}}}+{\left({2}{x}\right)}\frac{{\frac{{1}}{{4}}}}{{\frac{{x}}{{4}}}}+{1}$

$\displaystyle{y}''={2}{\ln{{\left(\frac{{x}}{{4}}\right)}}}+{3}$

$\displaystyle{2}{\ln{{\left(\frac{{x}}{{4}}\right)}}}+{3}={0}=\Rightarrow{\ln{{\left(\frac{{x}}{{4}}\right)}}}=-\frac{{3}}{{2}}=\Rightarrow{x}={4}{{e}}^{{-\frac{{3}}{{2}}}}\approx{.89}$ so there is an inflection point at $\displaystyle{x}={4}{{e}}^{{-\frac{{3}}{{2}}}}$ as the concavity changes from concave down to concave up.

The graph:

Ninenotes

Sunday, September 9, 2012

$\displaystyle{y}={{x}}^{{2}}{\ln{{\left(\frac{{x}}{{4}}\right)}}}$ Locate any relative extrema and points of inflection.

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