Answer
$$
f^{\prime \prime}(t)=\sqrt[3]{t}-\cos t=t^{1 / 3}-\cos t ,\quad f(0)=2 ,\quad f(1)=2
$$
The required function is
$$
f(t)=\frac{9}{28} t^{7 / 3}+\cos t+(\frac{19}{28}-cos (1)) t+1
$$
Work Step by Step
$$
f^{\prime \prime}(t)=\sqrt[3]{t}-\cos t=t^{1 / 3}-\cos t ,\quad f(0)=2 ,\quad f(1)=2
$$
The general anti-derivative of $
f^{\prime \prime}(x)=\sqrt[3]{t}-\cos t
$ is
$$
f^{\prime}(t)=\frac{3}{4} t^{4 / 3}-\sin t+C
$$
Using the anti-differentiation rules once more, we find that:
$$
f(t)=\frac{9}{28} t^{7 / 3}+\cos t+C t+D
$$
To determine $ C, D$ we use the fact that $f(0)=2, f(1)=2$:
$$
f(0)=\frac{9}{28} (0)^{7 / 3}+\cos (0)+C(0)+D=2
$$
$ \Rightarrow $
$$
1+D=2 \quad \Rightarrow \quad D=1,
$$
and
$$
f(1)=\frac{9}{28} (1)^{7 / 3}+\cos (1)+C (1)+1=2
$$
$ \Rightarrow $
$$
\frac{37}{28} +\cos (1)+C=2 \quad \Rightarrow \quad C=\frac{19}{28}-cos (1),
$$
so the required function is
$$
f(t)=\frac{9}{28} t^{7 / 3}+\cos t+(\frac{19}{28}-cos (1)) t+1
$$
