Problem 25 - Calculus of a Single Variable

Question

Finding Extrema on a Closed Interval In Exercises \(17-36\) , find the absolute extrema of the function on the closed interval. $$ g(t)=\frac{t^{2}}{t^{2}+3},[-1,1] $$

Step-by-Step Solution

Answer

\nThe absolute extrema of the function \( g(t) = \frac{t^{2}}{t^{2}+3} \) on the interval \([-1,1]\) are a maximum of \( \frac{1}{4} \) at \( t = -1 \) and \( t = 1 \), and a minimum of \( 0 \) at \( t = 0 \).

Step 1: Find the Derivative

First, find the derivative of the function \( g(t)=\frac{t^{2}}{t^{2}+3} \). Here we will need to use the quotient rule for differentiation which is \((f/g)'=(g'f-f'g)/g^2\). That gives us \( g'(t)=\frac{2t(t^{2}+3)-2t^{3}}{(t^{2}+3)^{2}} \). Simplifying the numerator we get \( g'(t)=\frac{6t}{(t^{2}+3)^{2}} \).

Step 2: Find the Critical Points

We set the derivative equal to zero and solve for 't' to find the critical points. This gives us \( 6t = 0 \), which implies that \( t = 0 \) is the only critical point.

Step 3: Evaluate the Function at Critical Points and Endpoints

Now, plug the critical points and the endpoints of the interval into the function to determine their output. Evaluating the function \( g(t) \) at \( t = -1, 0, 1 \) gives us \( g(-1) = \frac{1}{4}, g(0) = 0, g(1) = \frac{1}{4} \).

Step 4: Determine the Extrema

By comparing these outputs, we find that the absolute maximum of \( g(t) \) on the interval \([-1,1]\) is \( \frac{1}{4} \), and occurs at \( t = -1 \) and \( t = 1 \). The absolute minimum is \( 0 \), and only occurs at \( t = 0 \).

Key Concepts

Quotient Rule for Differentiation

The quotient rule for differentiation is a pivotal instrument in calculus when dealing with functions that consist of one function divided by another. If you have a function in the form of a quotient, such as \( g(t) = \frac{f(t)}{h(t)} \), you'll use this rule to differentiate it.

The quotient rule states that the derivative of \( \frac{f(t)}{h(t)} \) is given by:

• \((f/h)' = \frac{h'f - f'h}{h^2}\)

This rule requires the derivatives of both the numerator \(f(t)\) and the denominator \(h(t)\).

In our exercise, the function is \( g(t) = \frac{t^2}{t^2+3} \). Here, let \( f(t) = t^2 \) and \( h(t) = t^2 + 3 \). The derivatives are \( f'(t) = 2t \) and \( h'(t) = 2t \).

Applying the quotient rule, we compute:

• \( g'(t) = \frac{(2t)(t^2+3) - (t^2)(2t)}{(t^2+3)^2} \)
• Simplifying gives \( g'(t) = \frac{6t}{(t^2+3)^2} \)

The quotient rule's power lies in this ability to simplify and manage complex fractions efficiently.

Critical Points

Critical points are where a function's derivative is either zero or undefined. They play a key role in identifying where a function might have maximum or minimum values, which are essential for understanding the function's behavior.

To find critical points, take the derivative and set it equal to zero, then solve for the variable. In our case, with the derivative \( g'(t) = \frac{6t}{(t^2+3)^2} \), we set:

• \( 6t = 0 \)

Thus, the critical point is \( t = 0 \).

To further check if other forms like undefined points exist, consider the domain. Here, the denominator \((t^2 + 3)^2\) never equals zero, meaning it's always defined across all real numbers. Hence \( t = 0 \) is our sole critical point in the interval \([-1, 1]\).

Critical points are fundamental as they provide insight into the function's turning points, where these potential maximums and minimums may lie.

Absolute Maximum and Minimum

The absolute maximum and minimum of a function within a closed interval refer to the largest and smallest values the function attains over that interval. To find these points, evaluate the function at every critical point within the interval and at the interval's endpoints.

For our function \( g(t) = \frac{t^2}{t^2+3} \), we've identified the critical point \( t = 0 \). We also need to evaluate the function at the interval's endpoints \( t = -1 \) and \( t = 1 \).

• \( g(-1) = \frac{1}{4} \)
• \( g(0) = 0 \)
• \( g(1) = \frac{1}{4} \)

By comparing these values, the absolute maximum value on \([-1, 1]\) is \( \frac{1}{4} \) occurring at both \( t = -1 \) and \( t = 1 \).

The absolute minimum value is \( 0 \), which occurs at \( t = 0 \).

Such analysis is crucial, especially in optimization problems, where knowing the extremum values helps in decision-making or understanding phenomena within a specific boundary.