Problem 23 - Essential Calculus

Question

Locate 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

The absolute maximum value of the function on the interval [-1,1] is \( \frac{1}{4} \) at \( t = -1 \) and \( t = 1 \), and the absolute minimum value is 0 at \( t = 0 \).

Step 1: Calculate the derivative of the function

Differentiate the function \( g(t) \) with respect to \( t \) using the quotient rule for differentiation. The quotient rule is given as the derivative of \( \frac{u}{v} = \frac{vu' - uv'}{v^{2}} \), where \( u = t^{2} \) and \( v = t^{2} + 3 \). This gives \( g'(t) = \frac{2t(t^{2}+3) - t^{2}*2t}{(t^{2}+3)^2} = \frac{2t^{2}+6t - 2t^{3}}{(t^{2}+3)^2} \).

Step 2: Find the critical points of the function

To find the critical points of the function, set the derivative equal to zero and solve for \( t \). So, solving \( g'(t) = 0 \) yields \( t = 0 \).

Step 3: Evaluate the function at the critical points and endpoints

Evaluate the function at the critical point \( t = 0 \) and at the endpoints \( t = -1 \) and \( t = 1 \). This gives: \( g(-1) = \frac{1^2}{1^2+3} = \frac{1}{4} \), \( g(0) = \frac{0}{0+3} = 0 \), and \( g(1) = \frac{1^2}{1^2+3} = \frac{1}{4} \) .

Step 4: Identify the absolute extrema

The absolute maximum value of the function on the interval [-1,1] is \( \frac{1}{4} \) which occurs at \( t = -1 \) and \( t = 1 \), and the absolute minimum value is 0 at \( t = 0 \).

Key Concepts

Calculus

The branch of mathematics known as calculus is integral to analyzing change and motion.
It's divided into two main areas: differential calculus, which focuses on the concept of a derivative, and integral calculus, which is concerned with the concept of an integral.
The derivative measures how a function changes when its input changes, essentially describing the slope of the function at any point.
Solving problems involving absolute extrema is a common application of differential calculus.
These are the highest or lowest values that a function can take on a given interval and are essential in various fields, from engineering to economics.

Critical Points

In calculus, critical points are where the function's derivative is either zero or undefined.
These points are significant since they can indicate where a function attains its maximum or minimum values, which are the local extrema, or where the function's graph changes direction or has a point of inflection.
To find critical points, one typically sets the derivative of the function to zero and solves for the variable.
However, not all critical points will correspond to absolute extrema; it requires further analysis using methods like the First and Second Derivative Tests or Closed Interval Analysis to determine the nature of these points.

Quotient Rule for Differentiation

The Quotient Rule for Differentiation is a method for finding the derivative of a function that is the ratio of two differentiable functions.
The formula is \( \frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^{2}} \) where \( u '\) and \( v '\) are the derivatives of \( u \) and \( v \) with respect to \( x \), respectively.
It is important because it allows derivatives of complex ratios to be found without having to simplify the function first.

Closed Interval Analysis

To determine the absolute extrema of a function on a closed interval, one employs Closed Interval Analysis.
This procedure includes several critical steps:
First, find the derivative of the function and identify the critical points within the interval.
Second, evaluate the function at each critical point and at the endpoints of the interval.
Finally, compare these values to identify the smallest and largest ones, which are your absolute minimum and maximum, respectively.
This method ensures that you're considering all possible points where extrema could occur, including those at the boundaries of the interval which might be overlooked by considering only the critical points.