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Graphing utilities are incredibly useful tools for visualizing functions. They help us see the behavior of functions over specific intervals. In this case, we want to observe the function \( f(x) = (x^2 - 1) e^x \) on the interval \([-2, 2]\). By plotting the function using a graphing utility, you can identify where the function might have local or absolute maxima and minima. This visual approach provides a first impression of potential high and low points before diving into calculus for exact values.

• Graphing allows for easy spotting of trends and behavior.
• Visuals help in understanding where changes in direction occur.
• It provides a preliminary check for critical point calculations later on.

By first examining the graph, we can narrow down our focus to the most significant regions of interest for further analysis.

Critical points are locations on a curve where the derivative is either zero or undefined. These points are essential for determining possible local maxima and minima for a function within an interval. For the function \( f(x) = (x^2 - 1) e^x \), we first find the derivative and then set it equal to zero to find the critical points. The critical points we find provide candidate locations where the function could reach its highest or lowest values on the given interval.

• Critical points occur where \( f'(x) = 0 \) or is undefined.
• They are potential indicators for local extremum (minimums or maximums).
• Specific calculations give exact values, which the graph supports visually.

Accurate identification of these points is crucial in thoroughly understanding the behavior and limits of the function.

Calculating the derivative is a key step in locating critical points. For the function \( f(x) = (x^2 - 1) e^x \), we apply the product rule, since our function can be broken into \( g(x) = x^2 - 1 \) and \( h(x) = e^x \). The product rule states that \( (gh)' = g'h + gh' \). Applying this gives us:\[ f'(x) = (2x) e^x + (x^2 - 1) e^x \]\[ = (x^2 + 2x - 1) e^x \]By simplifying, the derivative expresses changes in the function, offering insights into the function’s slope at any point on the interval.

• The derivative connects the instantaneous rate of change with geometry.
• Applying rules correctly, like the product rule, ensures accurate results.
• Simplification reveals points of zero slope, helping find maxima and minima.

Understanding the derivative gives us a mathematical tool to confirm and explore our graph’s visual findings.

The quadratic formula is a vital algebraic tool used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). When finding critical points for \( f(x) = (x^2-1)e^x \), we end up solving the quadratic equation \( x^2 + 2x - 1 = 0 \). Here, the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) comes into play:For this function, we have \( a = 1 \), \( b = 2 \), and \( c = -1 \). Plug into the formula:\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2} \]These solutions \( x = -1 + \sqrt{2} \) and \( x = -1 - \sqrt{2} \) are the critical points needed for further function evaluation.

• Helps solve equations that are difficult to factor manually.
• Provides precise values for roots and critical points.
• Is a key step in transitioning from the derivative to exact function behavior.

The quadratic formula simplifies finding intersections or roots, supporting our calculus analysis by confirming the location of critical points.