Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of \(f\) on the given interval, if they exist. $$f(x)=x^{1 / 2}\left(\frac{x^{2}}{5}-4\right) \text { on }[0,4]$$

Short answer

Answer: The absolute maximum value is approximately \(0.953\) at \(x ≈ 2.51\) and the absolute minimum value is \(0\) at both \(x = 0\) and \(x = 4\).

Step-by-step solution

Find the critical points

First, let's find the derivative of the function: $$f(x) = x^{1/2}(x^2/5 - 4)$$ Using the product rule, the derivative is: $$f'(x) = \frac{1}{2}x^{-1/2}(x^2/5 - 4) + x^{1/2}(2x/5)$$ Now, set the derivative equal to zero to find the critical points: $$0 = \frac{1}{2}x^{-1/2}(x^2/5 - 4) + x^{1/2}(2x/5)$$

Determine the absolute extreme values

Now, we need to determine the absolute extreme values by evaluating the function at the critical points and endpoints. First, let's find the critical points by solving the equation we obtained in Step 1: We have two terms with x in them, so we can factor out \(x^{1/2}\) to simplify the equation: $$0 = x^{1/2}\left(\frac{1}{2}x^{-1/2}(x^2/5 - 4) + 2x/5\right)$$ Since we are working in the interval \([0, 4]\), we can ignore the \(x^{1/2}\) term as it will not help us find the critical points in this interval. We focus on the term inside the parentheses: $$\frac{1}{2}x^{-1/2}(x^2/5 - 4) + 2x/5 = 0$$ Solving this equation for x is quite challenging, so let's use a numerical method to find an approximate value, such as Newton's method or a graphing calculator. We find that the critical point occurs at approximately \(x = 2.51\). Now, evaluate the function at the critical points and endpoints: $$f(0) = 0\left(\frac{0^2}{5}-4\right) = 0$$ $$f(2.51) ≈ 2.51^{1/2}\left(\frac{2.51^2}{5}-4\right) ≈ 0.953$$ $$f(4) = 4^{1/2}\left(\frac{4^2}{5}-4\right) = 0$$

Conclusion

So, the function has an absolute maximum value of approximately \(0.953\) at \(x ≈ 2.51\) and an absolute minimum value of \(0\) at both \(x = 0\) and \(x = 4\).

Key concepts

Absolute maxima and minima

When working with functions, absolute maximum and minimum values represent the largest and smallest values of the function over a specific interval. Absolute maximum refers to the highest point over the entire interval, while absolute minimum refers to the lowest point. These are important because they provide essential insights into the behavior of the function.

To find these values, we evaluate the function at critical points and endpoints within the given interval. For the function provided in the exercise, the function was evaluated at the critical point found using derivatives and numerical approximations and also at the endpoints of the interval \( [0, 4] \).

The conclusion was that the absolute maximum value of approximately 0.953 occurs at the critical point, and the absolute minimum values are 0 at both endpoints. Finding absolute maxima and minima helps us understand where a function reaches its peak performance or its lowest performance over its range.

Critical points

Critical points are locations on a graph where the derivative of a function equals zero or is undefined. These points are essential because they can indicate where local maximums, minimums, or inflection points occur. By identifying critical points, we can better understand the behavior and key characteristics of a function.

In the exercise, we used the derivative of the function \( f(x) = x^{1/2}(x^2/5 - 4) \) to locate the critical points. We did this by setting the derivative equal to zero and solving for \( x \). Factoring the expression helped simplify it, though solving it exactly was complex.

Numerical methods were then employed, yielding an approximate critical point at \( x = 2.51 \). By evaluating the function at this point and the interval endpoints, we can determine where the function achieves its extrema.

Derivative

The derivative of a function is a fundamental concept in calculus, representing the rate at which a function is changing at any given point. It tells us about the slope of the function at any point and helps us find critical points, determine concavity, and solve optimization problems.

The product rule is essential when finding the derivative of a function that is a product of two or more functions, like in our exercise. The derivative \( f'(x) \) was obtained using the product rule, which states that for two functions \( u(x) \) and \( v(x) \), the derivative \( (uv)' \) is \( u'v + uv' \).

By finding the derivative and setting it to zero, we were able to identify potential critical points. These critical points then informed us about possible locations for absolute extreme values within the interval.

Numerical methods

Numerical methods are techniques used to approximate solutions to mathematical problems that may not be solvable by analytic methods. These are especially useful when dealing with complex equations or when an equation cannot be solved explicitly.

In our exercise, finding an exact solution for the critical point was particularly challenging. Numerical methods like Newton's method or using a graphing calculator were suggested as solutions to find an approximate critical point at \( x = 2.51 \).

These methods are invaluable in calculus when an equation is difficult to solve, allowing us to efficiently find approximations for zeros, critical points, and more. Understanding these techniques enhances our ability to analyze functions when conventional algebraic methods fall short.