If f ″ ( t ) = 3 t {\displaystyle \displaystyle f''(t)={\frac {3}{\sqrt {t}}}} , then antidifferentiating gives
f ′ ( t ) = 6 t + C {\displaystyle \displaystyle f'(t)=6{\sqrt {t}}+C} . Since f ′ ( 4 ) = 7 {\displaystyle \displaystyle f'(4)=7} , we have that
7 = f ′ ( 4 ) = 6 4 + C = 12 + C {\displaystyle \displaystyle 7=f'(4)=6{\sqrt {4}}+C=12+C}
and hence C=-5. Thus
f ′ ( t ) = 6 t − 5 {\displaystyle \displaystyle f'(t)=6{\sqrt {t}}-5}
and antidifferentiating once more gives
f ( t ) = 4 t 3 − 5 t + D {\displaystyle \displaystyle f(t)=4{\sqrt {t}}^{3}-5t+D}
Since f ( 4 ) = 2 {\displaystyle \displaystyle f(4)=2} , we see that
2 = f ( 4 ) = 4 4 3 − 5 ( 4 ) + D = 12 + D {\displaystyle \displaystyle 2=f(4)=4{\sqrt {4}}^{3}-5(4)+D=12+D}
and so D=-10. Thus, the function we seek is
f ( t ) = 4 t 3 − 5 t − 10 {\displaystyle \displaystyle f(t)=4{\sqrt {t}}^{3}-5t-10}
, then antidifferentiating gives
. Since 
, we have that
, we see that
