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# (a) Find the vertical and horizontal asymptotes.(b) Find the intervals of increase or decrease.(c) Find the local maximum and minimum values.(d) Find the intervals of concavity and the inflection points.(e) Use the information from parts $(a) - (d)$ to sketch the graph of $f$.$f(x) = x - \frac{1}{6}x^2 - \frac{2}{3} \ln x$

## a) Vertical asymptote is $x=0$There is no horizontal asymptoteb) Decreasing: $0 < x < 1$ $2 < x < \infty$ Increasing: $1< x < 2$c) if $x=1$ then $f(x)=1-\frac{1}{6} \times 1^{2}-\frac{2}{3} \ln 1=\frac{5}{6} \approx 0.833$if $x=2$ then $f(x)=2-\frac{1}{6} \times 2^{2}-\frac{2}{3} \ln 2 \approx 0.871$d) $(0, \sqrt{2})$ concave up$(\sqrt{2}, \infty)$ concave downInflection point $=\sqrt{2}$e) SEE GRAPH

Derivatives

Differentiation

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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Derivatives

Differentiation

Volume

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp