Example 3 Let the temperature $T$ of an object be a function of time $t$: $T(t)=a t^{3}+b t^{2}+c t+d(a \neq 0)$, where the unit of temperature is ${ }^{\circ} \mathrm{C}$, the unit of time is hours, and $t=0$ represents $12: 00$, with positive $t$ values indicating times after $12: 00$. If the temperature of the object is measured to be $8^{\circ} \mathrm{C}$ at $8: 00$, $60^{\circ} \mathrm{C}$ at $12: 00$, and $58^{\circ} \mathrm{C}$ at $13: 00$, and it is known that the rate of change of the object's temperature at $8: 00$ and $16: 00$ is the same.
(1) Write the function relating the object's temperature $T$ to time $t$;
(2) During the period from $10: 00$ to $14: 00$ (including $10: 00$ and $14: 00$), when is the temperature of the object the highest? And find the highest temperature.
(1) According to the conditions, we have $T(0)=60, T(-4)=8, T(1)=58, T^{\prime}(-4)=T^{\prime}(4)$, then $d=$ $60, b=0, a=1, c=-3$, therefore, the temperature function $T(t)=t^{3}-3 t+60$.
(2) $T^{\prime}(t)=3 t^{2}-3=3(t-1)(t+1)$, when $t \in(-2,-1) \cup(1,2)$, $T^{\prime}(t)>0$; when $t \in$ $(-1,1)$, $T^{\prime}(t)<0$. Therefore, the function $T(t)$ is increasing on $(-2,-1)$, decreasing on $(-1,1)$, and increasing on $(1,2)$, which means $t=-1$ is a point of local maximum. Since $T(-1)=T(2)=62$, the highest temperature of the object between $10:00$ and $14:00$ is $62^{\circ} \mathrm{C}$, occurring at $11:00$ and $14:00$.