Chapter 3 - Applications of Differentiation - 3.4 Exercises - Page 192: 24

$$\eqalign{ & {\text{Inflection point }}\left( {9,4} \right) \cr & {\text{Concave downward}}:{\text{ }}\left( {0,9} \right) \cr & {\text{Concave upward}}:{\text{ }}\left( {9,\infty } \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{x + 3}}{{\sqrt x }} \cr & {\text{The domain of the function is }}x > 0 \cr & {\text{Calculate the first and second derivatives}} \cr & f\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{x + 3}}{{\sqrt x }}} \right] \cr & f\left( x \right) = \frac{d}{{dx}}\left[ {\frac{x}{{\sqrt x }} + \frac{3}{{\sqrt x }}} \right] \cr & f\left( x \right) = \frac{d}{{dx}}\left[ {{x^{1/2}} + 3{x^{ - 1/2}}} \right] \cr & f'\left( x \right) = \frac{1}{2}{x^{ - 1/2}} - \frac{3}{2}{x^{ - 3/2}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{2}{x^{ - 1/2}} - \frac{3}{2}{x^{ - 3/2}}} \right] \cr & f''\left( x \right) = \frac{1}{2}\left( { - \frac{1}{2}} \right){x^{ - 1/2 - 1}} - \frac{3}{2}\left( { - \frac{3}{2}} \right){x^{ - 3/2 - 1}} \cr & f''\left( x \right) = - \frac{1}{4}{x^{ - 3/2}} + \frac{9}{4}{x^{ - 5/2}} \cr & {\text{Set the second derivative to }}0 \cr & - \frac{1}{4}{x^{ - 3/2}} + \frac{9}{4}{x^{ - 5/2}} = 0 \cr & - \frac{1}{4}{x^{ - 5/2}}\left( {x - 9} \right) = 0 \cr & - \frac{1}{{4{x^{5/2}}}}\left( {x - 9} \right) = 0 \cr & x = 9 \cr & {\text{The second derivative is not defined at }}x = 0 \cr & {\text{The value }}x = 0{\text{ is not in the domain of the function }} \cr & {\text{Set the intervals }}\left( {0,9} \right),\left( {9,\infty } \right) \cr & {\text{Making a table of values }}\left( {{\text{See examples on page 188 }}} \right) \cr} $$ \[\boxed{\begin{array}{*{20}{c}} {{\text{Interval}}}&{\left( {0,9} \right)}&{\left( {9,\infty } \right)} \\ {{\text{Test Value}}}&{x = 1}&{x = 16} \\ {{\text{Sign of }}f''\left( x \right)}&{f''\left( 1 \right) = 2 > 0}&{f''\left( 5 \right) = - \frac{7}{{4096}} < 0} \\ {{\text{Conclusion}}}&{{\text{Concave downward}}}&{{\text{Concave downward}}} \end{array}}\] $$\eqalign{ & {\text{Inflection point }}\left( {9,4} \right) \cr & {\text{Concave downward}}:{\text{ }}\left( {0,9} \right) \cr & {\text{Concave upward}}:{\text{ }}\left( {9,\infty } \right) \cr} $$