#### Answer

$$f''\left( x \right) = 2{x^2}{e^{{x^2}}} + {e^{{x^2}}},f''\left( 0 \right) = 1,f''\left( 2 \right) = 9{e^4}$$

#### Work Step by Step

$$\eqalign{
& f\left( x \right) = 0.5{e^{{x^2}}} \cr
& {\text{find the derivative of }}f\left( x \right) \cr
& {\text{ }}f'\left( x \right) = \frac{d}{{dx}}\left[ {0.5{e^{{x^2}}}} \right] \cr
& {\text{ }}f'\left( x \right) = 0.5\frac{d}{{dx}}\left[ {{e^{{x^2}}}} \right] \cr
& {\text{use rule }}\frac{d}{{dx}}\left[ {{e^{g\left( x \right)}}} \right] = {e^{g\left( x \right)}}g'\left( x \right).{\text{ set }}g\left( x \right) = {x^2} \cr
& {\text{ }}f'\left( x \right) = 0.5{e^{{x^2}}}\frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& {\text{ }}f'\left( x \right) = 0.5{e^{{x^2}}}\left( {2x} \right) \cr
& {\text{ }}f'\left( x \right) = x{e^{{x^2}}} \cr
& \cr
& {\text{find the derivative of }}f'\left( x \right) \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {x{e^{{x^2}}}} \right] \cr
& {\text{by using the product rule}} \cr
& f''\left( x \right) = x\frac{d}{{dx}}\left[ {{e^{{x^2}}}} \right] + {e^{{x^2}}}\frac{d}{{dx}}\left[ x \right] \cr
& f''\left( x \right) = x\left( {2x{e^{{x^2}}}} \right) + {e^{{x^2}}}\left( 1 \right) \cr
& {\text{simplifying}} \cr
& f''\left( x \right) = 2{x^2}{e^{{x^2}}} + {e^{{x^2}}} \cr
& \cr
& {\text{find }}f''\left( 0 \right){\text{ and }}f''\left( 2 \right) \cr
& f''\left( 0 \right) = 2{\left( 0 \right)^2}{e^{{{\left( 0 \right)}^2}}} + {e^{{{\left( 0 \right)}^2}}} = 0 + 1 = \cr
& f''\left( 2 \right) = 2{\left( 2 \right)^2}{e^{{{\left( 2 \right)}^2}}} + {e^{{{\left( 2 \right)}^2}}} = 8{e^4} + {e^4} = 9{e^4} \cr} $$