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 temperature is in ${ }^{\circ} \mathrm{C}$ and the time is in hours, with $t=0$ representing 12:00, and positive $t$ values representing 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) At what time between 10:00 and 14:00 (inclusive) 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 $14:00$.