Let's solve the given problem step-by-step.

Problem Recap:

We are asked to find the exact global maximum and minimum values for the function: $f(t) = \frac{5t}{3 + t^2}$ over all real numbers $t$.

Step 1: Find the derivative $f'(t)$

We use the quotient rule for differentiation: $\frac{d}{dt} \left( \frac{u(t)}{v(t)} \right) = \frac{u'(t) v(t) - u(t) v'(t)}{[v(t)]^2}$ where $u(t) = 5t$ and $v(t) = 3 + t^2$.

1. $u'(t) = 5$
2. $v'(t) = 2t$

Now, apply the quotient rule: $f'(t) = \frac{5(3 + t^2) - 5t(2t)}{(3 + t^2)^2}$ Simplify the numerator: $5(3 + t^2) - 10t^2 = 15 + 5t^2 - 10t^2 = 15 - 5t^2$ So: $f'(t) = \frac{15 - 5t^2}{(3 + t^2)^2}$

Step 2: Find critical points by setting $f'(t) = 0$

We need to solve: $15 - 5t^2 = 0$ $5t^2 = 15 \quad \Rightarrow \quad t^2 = 3 \quad \Rightarrow \quad t = \pm \sqrt{3}$ So, the critical points are $t = \sqrt{3}$ and $t = -\sqrt{3}$.

Step 3: Evaluate $f(t)$ at the critical points and at the limits $t \to \pm \infty$

1. At $t = \sqrt{3}$: $f(\sqrt{3}) = \frac{5\sqrt{3}}{3 + (\sqrt{3})^2} = \frac{5\sqrt{3}}{3 + 3} = \frac{5\sqrt{3}}{6}$

2. At $t = -\sqrt{3}$: $f(-\sqrt{3}) = \frac{5(-\sqrt{3})}{3 + (\sqrt{3})^2} = \frac{-5\sqrt{3}}{6}$

3. As $t \to \pm \infty$: $\lim_{t \to \pm \infty} f(t) = \lim_{t \to \pm \infty} \frac{5t}{3 + t^2} = 0$ So, the function approaches 0 as $t$ grows very large or very small.

Step 4: Determine the global maximum and minimum

• $f(\sqrt{3}) = \frac{5\sqrt{3}}{6}$ is the global maximum.
• $f(-\sqrt{3}) = -\frac{5\sqrt{3}}{6}$ is the global minimum.

Final Answer

• Global maximum at $t = \sqrt{3}$: $\frac{5\sqrt{3}}{6}$.
• Global minimum at $t = -\sqrt{3}$: $-\frac{5\sqrt{3}}{6}$.

Do you have any further questions or need more details?

Here are five related questions to explore:

1. How does the quotient rule for derivatives work?
2. What is the significance of critical points in calculus?
3. Why do we evaluate limits at $t \to \infty$ for global extrema?
4. How do you know if a critical point is a maximum or minimum?
5. Can the global extrema change if the domain is restricted?

Tip: To confirm extrema, you could also use the second derivative test.