Question

a. Find the critical points of the following functions on the given interval. b. Use a graphing utility to determine whether the critical points correspond to local maxima, local minima, or neither. c. Find the absolute maximum and minimum values on the given interval when they exist. $$f(t)=3 t /\left(t^{2}+1\right) \text { on }[-2,2]$$

Short answer

Answer: The absolute maximum value is \(f(1) = \frac{3}{2}\) and the absolute minimum value is \(f(-1) = -\frac{3}{2}\).

Step-by-step solution

Find the critical points

To find the critical points, we have to find the first derivative of the function and then determine where the derivative is equal to zero or does not exist. Given function: $$f(t) = \frac{3t}{t^2 + 1}$$. First, find the derivative of the function with respect to t: $$f'(t) = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{3t}{t^2 + 1}\right)$$ We use the quotient rule: \((u/v)' = (vu' - uv')/v^2\), where \(u = 3t\) and \(v = t^2 + 1\). Find the derivatives of u and v with respect to t: $$u'(t) = \frac{\mathrm{d}}{\mathrm{d}t}(3t) = 3$$ $$v'(t) = \frac{\mathrm{d}}{\mathrm{d}t}(t^2 + 1) = 2t$$ Now, apply the quotient rule: $$f'(t) = \frac{(t^2 + 1) \cdot 3 - 3t \cdot 2t}{(t^2 + 1)^2} = \frac{3t^2 + 3 - 6t^2}{(t^2 + 1)^2} = \frac{-3t^2 + 3}{(t^2 + 1)^2}$$ To find the critical points, we need to find where the derivative is 0 or undefined. $$f'(t) = \frac{-3t^2 + 3}{(t^2 + 1)^2} = 0$$

Determine the type of critical points

To determine the type of the critical points (maxima, minima, or neither), we can either use the second derivative test or, as asked in the exercise, use a graphing utility. By using a graphing utility or software (such as Desmos or Geogebra) to plot the function: $$f(t) = \frac{3t}{t^2 + 1}$$ We can observe that there is a local maximum around \(t = -1\) and a local minimum around \(t = 1\).

Find the absolute maximum and minimum values on the given interval

To find the absolute maximum and minimum values on the interval [-2,2], we need to evaluate the function at the critical points and endpoints of the interval. Then, compare the values. Critical points: \(t \approx -1\) and \(t \approx 1\) Endpoints: \(t = -2\) and \(t = 2\) Evaluate the function at these points: $$f(-1) = \frac{3(-1)}{(-1)^2 + 1} = -\frac{3}{2}$$ $$f(1) = \frac{3(1)}{(1)^2 + 1} = \frac{3}{2}$$ $$f(-2) = \frac{3(-2)}{(-2)^2 + 1} = -\frac{6}{5}$$ $$f(2) = \frac{3(2)}{(2)^2 + 1} = \frac{6}{5}$$ By comparing these values, we find that: - The absolute maximum value is \(f(1) = \frac{3}{2}\), at \(t = 1\). - The absolute minimum value is \(f(-1) = -\frac{3}{2}\), at \(t = -1\).

Key concepts

First Derivative

Understanding the first derivative of a function is crucial to finding its critical points. In calculus, the first derivative represents the rate of change or slope of the function at any given point. For a function like \( f(t) = \frac{3t}{t^2 + 1} \), the first derivative is determined by applying the derivative rules. This involves the quotient rule, where if \( u = 3t \) and \( v = t^2 + 1 \), the derivative \( f'(t) \) becomes:

• \( f'(t) = \frac{(v \cdot u' - u \cdot v')}{v^2} \)
• \( u'(t) = 3 \) and \( v'(t) = 2t \)
• Thus, \( f'(t) = \frac{3t^2 + 3 - 6t^2}{(t^2 + 1)^2} = \frac{-3t^2 + 3}{(t^2 + 1)^2} \)

The critical points occur where this derivative is zero or undefined. In this case, setting the derivative equal to zero allows us to solve for \( t \) values where changes in slope indicate potential maxima or minima. Critical points help identify specific turning points in the function's behavior within a given interval.

Derivative Test

Once the first derivative is known, the next step is to determine the nature of the critical points using the derivative test. This can involve either the first derivative test or the second derivative test. However, in this scenario, a graphing utility is favored.Using the first derivative test involves examining the sign changes of \( f'(t) \). If \( f'(t) \) changes from negative to positive at a critical point, it indicates a local minimum. Conversely, a change from positive to negative represents a local maximum. If the first derivative does not change sign, the critical point might be neither a maximum nor a minimum.For this exercise, observing the graph from a graphing utility simplifies the process. Visual representation shows how \( f(t) \) behaves around the critical points, providing a clear visual of local maxima and minima. Graphing verifies that around \( t = -1 \) there is a local maximum, and around \( t = 1 \) there is a local minimum.

Graphing Utility

Graphing utilities are powerful tools for visualizing the behavior of functions over an interval, allowing us to observe critical points instantly. By using software like Desmos or GeoGebra, you can plot the function \( f(t) = \frac{3t}{t^2 + 1} \) directly and visually assess where the local maxima and minima occur.To utilize a graphing utility:

• Enter the function into the graphing calculator.
• Adjust the interval to be \([-2, 2]\) as specified in the exercise.
• Zoom in to interactively see where the function's slope changes direction, which denotes critical points.

Seeing these points on a graph provides an immediate understanding of where and what type of extrema occur. The visual clarity from graphing utilities supplements analytical methods, confirming results obtained from calculus tests.

Absolute Maximum and Minimum

Finding the absolute maximum and minimum involves evaluating the function at the critical points and the endpoints of the given interval. This ensures you have accounted for all possible extremities within the interval.For \( f(t) = \frac{3t}{t^2 + 1} \), the critical points are \( t = -1 \) and \( t = 1 \), both derived from setting the first derivative equal to zero. The endpoints given are \( t = -2 \) and \( t = 2 \).Evaluate the function at these points:

• \( f(-1) = -\frac{3}{2} \), which is the absolute minimum.
• \( f(1) = \frac{3}{2} \), which is the absolute maximum.
• Endpoints: \( f(-2) = -\frac{6}{5} \) and \( f(2) = \frac{6}{5} \).

Comparing these values shows that the highest value at \( t = 1 \) makes it the absolute maximum, while the lowest at \( t = -1 \) represents the absolute minimum for the interval \([-2, 2]\). Evaluating both critical points and endpoints ensures no extreme values are overlooked.