Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of \(f\) on the given interval, if they exist. $$f(x)=x^{2}-10 \text { on }[-2,3]$$

Short answer

Answer: The absolute maximum value is -1 and occurs at x=3, and the absolute minimum value is -10 and occurs at x=0.

Step-by-step solution

Find the critical points

First, we need to find the derivative of the function and then set it equal to zero to find the critical points. The derivative of \(f(x) = x^2 - 10\) is $$f'(x) = 2x$$ Now, set the derivative equal to zero and solve for x: $$2x = 0 \implies x = 0$$ So, there is one critical point, \(x = 0\), within the interval \([-2,3]\).

Evaluate the function at critical points and endpoints

Now, we will evaluate the function at the critical point (\(x=0\)) and the endpoints of the interval (\(x=-2\) and \(x=3\)): $$f(0) = (0)^2 - 10 = -10$$ $$f(-2) = (-2)^2 - 10 = -6$$ $$f(3) = (3)^2 - 10 = 9 - 10 = -1$$

Identify the absolute maximum and minimum values

Comparing the function values, we can identify the absolute maximum and minimum values and their corresponding x-values: Absolute maximum value: \(f(3) = -1\) Absolute minimum value: \(f(0) = -10\) So, the absolute maximum value of \(f(x) = x^2 - 10\) on the interval \([-2,3]\) is \(-1\) and occurs at \(x=3\). The absolute minimum value is \(-10\) and occurs at \(x=0\).

Key concepts

Critical Points in Calculus

In calculus, critical points are spots on the graph of a function where the derivative is zero or undefined. These points are critical for understanding the overall behaviour of the function because they can indicate where the function might have relative extrema (local highs or lows).

To find critical points, one must take the derivative of the function and solve for where that derivative equals zero. In the given exercise, this process revealed that the function f(x) = x^2 - 10 has a critical point at x = 0 because its derivative f'(x) = 2x, when set equal to zero, leads to the solution x = 0. Thus, critical points are essential for determining where to evaluate a function to find absolute extrema on a given interval.

Function Evaluation

The concept of function evaluation is fundamental in understanding the behaviour of functions at specific points. It involves substituting a particular value into a function to find the output or the y-value. This is crucial when searching for extreme values on a closed interval.

For the exercise in question, function evaluation is used at the critical point x = 0, and at the boundary points of the interval, x = -2 and x = 3. By plugging these x-values into the f(x), we are able to compare the function values: f(0) = -10, f(-2) = -6, and f(3) = -1. This comparison allows us to establish which of these y-values are absolute maximums or minimums within the interval. Proper evaluation of functions is a key step in solving many problems in calculus.

Derivative of a Function

The derivative of a function is the rate at which the function's value changes at a given point. It's a fundamental concept in calculus because it gives insight into the function's behaviour, such as identifying increasing or decreasing intervals and finding critical points that could lead to extrema.

In the provided example, finding the derivative was the first step to unveiling where the function could possibly have extreme values. The derivative f'(x) = 2x led to the discovery of a critical point where x = 0. By analyzing the derivative, students gain the ability to predict the behaviour of functions, calculate slopes of tangent lines at any point, and solve a variety of practical problems in physics, engineering, economics, and beyond.