# Find f'(t) and f(t). f"(t)=t-cos t, f'(0)=6, f(0)=-6

Find f'(t) and f(t).
f"(t)=t-\cos t, f'(0)=6, f(0)=-6
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Joel Reese
$f"\left(t\right)=t-\mathrm{cos}t,{f}^{\prime }\left(0\right)=6,f\left(0\right)=-6$
Integrating both sicles, we have
${f}^{\prime }\left(t\right)=\frac{{t}^{2}}{2}-\mathrm{sin}t+{c}_{1}$
As f'(0)=6
$⇒6=0-\mathrm{sin}0+{c}_{1}⇒{c}_{1}=6$
$⇒{f}^{\prime }\left(t\right)=\frac{{t}^{2}}{2}-\mathrm{sin}t+6$
Again Integrating wrt t, we have
$⇒f\left(t\right)=\frac{1}{2}\frac{{t}^{3}}{3}+\mathrm{cos}t+6t+{c}_{2}$
As f(0)=-6
$⇒-6=0+\mathrm{cos}0+0+{c}_{2}$
$⇒-6=1+{c}_{2}⇒{c}_{2}=-7$
$⇒f\left(t\right)=\frac{{t}^{3}}{6}+\mathrm{cos}t+6t-7$
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