Answer
$$
f^{\prime \prime}(t)=4-\frac{6}{t^{4}} ,\quad f(1)=6 ,\quad f^{\prime}(2)=9,\quad t\gt 0
$$
The required function is
$$
f(t)=2t^{2} -t^{-2}+\frac{3}{4}t+\frac{17}{4}
$$
Work Step by Step
$$
f^{\prime \prime}(t)=4-\frac{6}{t^{4}} ,\quad f(1)=6 ,\quad f^{\prime}(2)=9,\quad t\gt 0
$$
can be written as the following:
$$
f^{\prime \prime}(t)=4-\frac{6}{t^{4}}=4-6t^{-4}
$$
The general anti-derivative of $
f^{\prime \prime}(t)=4-\frac{6}{t^{4}}
$ is
$$
f^{\prime}(t)=4t +2t^{-3}+C
$$
To determine C we use the fact that $f^{\prime}(2)=9$:
$$
f^{\prime}(2)=4(2) +2(2)^{-3}+C =9
$$
$ \Rightarrow $
$$
\frac{33}{4}+C=9\quad \Rightarrow \quad C=9-\frac{33}{4}=\frac{3}{4},
$$
so
$$
f^{\prime}(t)=4t +2t^{-3}+\frac{3}{4}
$$
Using the anti-differentiation rules once more, we find that:
$$
f(t)=2t^{2} -t^{-2}+\frac{3}{4}t+D
$$
To determine D we use the fact that $f(1)=6 $:
$$
f(1)=2(1)^{2} -(1)^{-2}+\frac{3}{4}(1)+D=6
$$
$ \Rightarrow $
$$
\frac{7}{4}+D=6\quad \Rightarrow \quad D=\frac{17}{4},
$$
so the required function is
$$
f(t)=2t^{2} -t^{-2}+\frac{3}{4}t+\frac{17}{4}
$$
