The solution shown in the image is finding the extreme time $t$ where the temperature $T(t)$ of a laboratory reaches a maximum or minimum. The method used is the Second Derivative Test. Here’s a step-by-step explanation:

1. Given Function:

The temperature $T(t)$ is modeled as: $T(t) = 42.5 - 17.4t + 3.57t^2 - 0.987t^3 + 0.0448t^4$

We aim to find $t$ at which the temperature is at a maximum or minimum.

2. Find the First Derivative $T'(t)$:

The first derivative gives the rate of change of temperature with respect to time $t$. Differentiating $T(t)$: $T'(t) = -17.4 + 7.14t - 2.961t^2 + 0.1792t^3$

To find stationary points (potential maximum or minimum points), set $T'(t) = 0$.

3. Solve $T'(t) = 0$:

$-17.4 + 7.14t - 2.961t^2 + 0.1792t^3 = 0$

Using numerical or analytical methods, the solutions are: $t = 14.12, \quad t = 1.16 + 2.34i, \quad t = 1.16 - 2.34i$

Since $t$ must be a real value (time cannot be complex), only $t = 14.12$ is valid.

4. Find the Second Derivative $T''(t)$:

The second derivative helps determine the nature of the stationary point. Differentiating $T'(t)$: $T''(t) = 7.14 - 5.922t + 0.5376t^2$

5. Evaluate $T''(t)$ at $t = 14.12$:

Substitute $t = 14.12$ into $T''(t)$: $T''(14.12) = 7.14 - 5.922(14.12) + 0.5376(14.12)^2$

After calculation: $T''(14.12) = 31.21 > 0$

Since $T''(t) > 0$, the stationary point at $t = 14.12$ corresponds to a minimum temperature.

6. Conclusion:

The temperature reaches a minimum at $t = 14.12$.

Do you need details about specific steps, or want me to clarify any part?

Related Questions:

1. What happens if $T''(t) < 0$ at a stationary point?
2. How do you solve a cubic equation like $T'(t) = 0$?
3. Could a second derivative test fail? What should be done in such cases?
4. Why are complex roots discarded in real-world problems like this one?
5. What is the physical significance of finding a minimum temperature?

Tip:

For polynomial functions, graphing can provide a visual understanding of where the maxima and minima lie!