Chapter 3 - Applications of Differentiation - Review Exercises - Page 238: 37

$$\eqalign{ & {\text{Concave upward}}:{\text{ }}\left( { - 5,\infty } \right) \cr & {\text{No point of inflection}} \cr} $$

$$\eqalign{ & g\left( x \right) = x\sqrt {x + 5} \cr & {\text{The domain of the function is }}x + 5 \geqslant 0 \cr & x \geqslant - 5 \cr & {\text{Calculate the first and second derivatives}} \cr & g'\left( x \right) = \frac{d}{{dx}}\left[ {x\sqrt {x + 5} } \right] \cr & g'\left( x \right) = \sqrt {x + 5} \frac{d}{{dx}}\left[ x \right] + x\frac{d}{{dx}}\left[ {\sqrt {x + 5} } \right] \cr & g'\left( x \right) = \sqrt {x + 5} + x\left( {\frac{1}{{2\sqrt {x + 5} }}} \right) \cr & g'\left( x \right) = \sqrt {x + 5} + \frac{x}{{2\sqrt {x + 5} }} \cr & g'\left( x \right) = \frac{{2\left( {x + 5} \right) + x}}{{2\sqrt {x + 5} }} \cr & g'\left( x \right) = \frac{{2x + 10 + x}}{{2\sqrt {x + 5} }} \cr & g''\left( x \right) = \underbrace {\frac{d}{{dx}}\left[ {\frac{{3x + 10}}{{2\sqrt {x + 5} }}} \right]}_{{\text{Use quotient rule}}} \cr & g''\left( x \right) = \frac{{2\sqrt {x + 5} \left( 3 \right) - \left( {3x + 10} \right)\left( {\frac{1}{{\sqrt {x + 5} }}} \right)}}{{{{\left( {2\sqrt {x + 3} } \right)}^2}}} \cr & g''\left( x \right) = \frac{{6\sqrt {x + 5} - \left( {3x + 10} \right)\left( {\frac{1}{{\sqrt {x + 5} }}} \right)}}{{{{\left( {2\sqrt {x + 5} } \right)}^2}}} \cr & g''\left( x \right) = \frac{{6\left( {x + 5} \right) - \left( {3x + 10} \right)\left( 1 \right)}}{{4{{\left( {x + 5} \right)}^{3/2}}}} \cr & g''\left( x \right) = \frac{{6x + 30 - 3x - 10}}{{4{{\left( {x + 5} \right)}^{3/2}}}} \cr & g''\left( x \right) = \frac{{3x + 20}}{{4{{\left( {x + 5} \right)}^{3/2}}}} \cr & {\text{Set the second derivative to }}0 \cr & \frac{{3x + 20}}{{4{{\left( {x + 5} \right)}^{3/2}}}} = 0 \cr & 3x + 20 = 0 \cr & x = - \frac{{20}}{3} \approx - 6.666 \cr & {\text{This value is not in the domain of the function }}g\left( x \right) = x\sqrt {x + 5} \cr & {\text{So, there are no inflection points, and the domain is }}x \geqslant - 5 \cr & {\text{Evaluating the second derivative at the interval }}\left( { - 5,\infty } \right) \cr} $$ \[\boxed{\begin{array}{*{20}{c}} {{\text{Interval}}}&{\left( { - 5,\infty } \right)} \\ {{\text{Test Value}}}&{x = - 4} \\ {{\text{Sign of }}g''\left( x \right)}&{g''\left( { - 4} \right) = 2 > 0} \\ {{\text{Conclusion}}}&{{\text{Concave upward}}} \end{array}}\] $$\eqalign{ & {\text{There are no inflection points}} \cr & {\text{Concave upward}}:{\text{ }}\left( { - 5,\infty } \right) \cr} $$