## Calculus 8th Edition

$$f^{\prime \prime}(t)=\sqrt[3]{t}-\cos t=t^{1 / 3}-\cos t ,\quad f(0)=2 ,\quad f(1)=2$$ The required function is $$f(t)=\frac{9}{28} t^{7 / 3}+\cos t+(\frac{19}{28}-cos (1)) t+1$$
$$f^{\prime \prime}(t)=\sqrt[3]{t}-\cos t=t^{1 / 3}-\cos t ,\quad f(0)=2 ,\quad f(1)=2$$ 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 $$f(t)=\frac{9}{28} t^{7 / 3}+\cos t+(\frac{19}{28}-cos (1)) t+1$$