Question

In Exercises, find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results. $$ g(t)=\frac{t^{2}}{t^{2}+3}, \quad[-1,1] $$

Short answer

The absolute minimum of the function is 0 at \(t = 0\) and the absolute maximum is \(\frac{1}{4}\) at \(t = -1\) and \(t = 1\).

Step-by-step solution

Identify the Function and Its Derivative

The given function is \(g(t) = \frac{t^2}{t^2 + 3}\). Calculate its derivative, \(g'(t)\) using the quotient rule, which gives \(g'(t) = \frac{2t(t^2 + 3) - 2t^3}{(t^2 + 3)^2}\).\nThis simplifies to \(g'(t) = \frac{6t}{(t^2 + 3)^2}\).

Find the Critical Points

To find the critical points, set the derivative \( g'(t)\) equal to zero and solve for \(t\). From \( g'(t) = \frac{6t}{(t^2 + 3)^2} = 0\), we find that \(t = 0\) is the only critical point in the interval [-1, 1].

Evaluate the Function at critical and boundary points

We evaluate the function at \(t = 0\) (the critical point) and at \(t = -1\) and \(t = 1\) (the boundaries of the interval). We find that \(g(0) = 0\), \(g(-1) = \frac{1}{4}\), and \(g(1) = \frac{1}{4}\).

Identify the Absolute Extrema

The smallest value \(g(t)\) takes on is \(g(0) = 0\), and thus the absolute minimum is 0 at \(t = 0\). The largest value \(g(t)\) takes on is \(g(-1) = \frac{1}{4}\) or \(g(1) = \frac{1}{4}\), so the absolute maximum is \(\frac{1}{4}\) at \(t = -1\) and \(t = 1\). It is important to validate these calculations with a graphing utility.

Key concepts

Critical Points

In the quest to determine the highs and lows of a function within a certain interval, the concept of critical points is central. Simply put, critical points occur where the derivative of the function is either zero or is undefined. These points are potential contenders for where a function may achieve its maximum or minimum values.

To find critical points, one must first understand derivatives. A derivative essentially represents the rate at which the function is changing at any given point. In the context of the function g(t) = \( \frac{t^2}{t^2 + 3} \), we look for points in the domain where the derivative g'(t) equals zero. After applying the necessary calculus tools, we identify t = 0 as our sole critical point within the interval [-1, 1].

This critical point, along with the boundary points, are examined to determine the absolute extrema. Therefore, knowing how to properly identify and interpret critical points is pivotal for analyzing the behavior of functions and ultimately understanding calculus.

Quotient Rule

When faced with the task of differentiating a function that consists of one function divided by another, the quotient rule is a reliable ally. It states that if you have a function that can be represented as f(t)/g(t), where both f(t) and g(t) are themselves differentiable functions, the derivative of the whole is (g(t)*f'(t) - f(t)*g'(t))/(g(t))^2.

Applying the quotient rule to our given function g(t) = \( \frac{t^2}{t^2 + 3} \), we differentiate the top and bottom separately to obtain g'(t) = \( \frac{2t(t^2 + 3) - 2t^3}{(t^2 + 3)^2} \), which simplifies to g'(t) = \( \frac{6t}{(t^2 + 3)^2} \). Clear understanding and correct application of the quotient rule are essential in calculus, especially when we're trying to locate critical points where the value of the derivative is zero.

Thus, mastering this rule not only simplifies complex differentiation but is the stepping stone to finding the extrema of functions involving ratios.

Graphing Utility

Modern mathematics greatly benefits from technology, especially when it comes to visualizing concepts and verifying results like absolute extrema. The use of a graphing utility allows students and mathematicians alike to draw precise graphs of functions with minimal effort.

By inputting our function g(t) = \( \frac{t^2}{t^2 + 3} \) into a graphing utility, we can observe the curve and confirm the location of our absolute maximum and minimum found analytically. For this function, seeing the curve on the interval [-1, 1] reaffirms that at t = 0, the function indeed dips to its lowest value, while at the boundaries t = -1 and t = 1, it peaks at its highest.

The visual confirmation through graphing utilities can provide reassurance that the algebraic procedures have been conducted correctly, making it a valuable tool in a mathematician's toolkit. Whether it's a complex function or simply verifying the behavior at critical points, a graphing utility can illuminate the path to a better understanding of mathematical functions.