## Calculus with Applications (10th Edition)

$${\text{ }}f''\left( x \right) = \frac{{112}}{{{{\left( {4x + 5} \right)}^3}}},\,\,\,\,\,f''\left( 1 \right) = \frac{{112}}{{729}},\,\,\,\,f''\left( { - 3} \right) = - \frac{{16}}{{49}}$$
\eqalign{ & f\left( x \right) = \frac{{1 - 2x}}{{4x + 5}} \cr & {\text{find the derivative of }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{1 - 2x}}{{4x + 5}}} \right] \cr & {\text{by using the quotient rule }}\frac{d}{{dx}}\left[ {\frac{{u\left( x \right)}}{{v\left( x \right)}}} \right] = \frac{{v\left( x \right) \cdot u'\left( x \right) - u\left( x \right)v'\left( x \right)}}{{{{\left[ {v\left( x \right)} \right]}^2}}} \cr & f'\left( x \right) = \frac{{\left( {4x + 5} \right)\frac{d}{{dx}}\left[ {1 - 2x} \right] - \left( {1 - 2x} \right)\frac{d}{{dx}}\left[ {4x + 5} \right]}}{{{{\left( {4x + 5} \right)}^2}}} \cr & {\text{solve derivatives}} \cr & f'\left( x \right) = \frac{{\left( {4x + 5} \right)\left( { - 2} \right) - \left( {1 - 2x} \right)\left( 4 \right)}}{{{{\left( {4x + 5} \right)}^2}}} \cr & {\text{simplifying}} \cr & f'\left( x \right) = \frac{{ - 8x - 10 - 4 + 8x}}{{{{\left( {4x + 5} \right)}^2}}} \cr & f'\left( x \right) = \frac{{ - 14}}{{{{\left( {4x + 5} \right)}^2}}} \cr & {\text{by using the quotient rule}} \cr & f''\left( x \right) = \frac{{{{\left( {4x + 5} \right)}^2}\frac{d}{{dx}}\left( { - 14} \right) + 14\frac{d}{{dx}}\left[ {{{\left( {4x + 5} \right)}^2}} \right]}}{{{{\left( {{{\left( {4x + 5} \right)}^2}} \right)}^2}}} \cr & f''\left( x \right) = \frac{{14\left( 2 \right)\left( {4x + 5} \right)\left( 4 \right)}}{{{{\left( {{{\left( {4x + 5} \right)}^2}} \right)}^2}}} \cr & {\text{simplifying}} \cr & f''\left( x \right) = \frac{{112}}{{{{\left( {4x + 5} \right)}^3}}} \cr & \cr & {\text{find }}f''\left( 1 \right){\text{ and }}f''\left( { - 3} \right) \cr & f''\left( 1 \right) = \frac{{112}}{{{{\left( {4\left( 1 \right) + 5} \right)}^3}}} = \frac{{112}}{{729}} \cr & and \cr & f''\left( { - 3} \right) = \frac{{112}}{{{{\left( {4\left( { - 3} \right) + 5} \right)}^3}}} = - \frac{{16}}{{49}} \cr}