Problem 33 - Applied Calculus

Question

(a) For \(a\) a positive constant, find all critical points of \(f(x)=x-a \sqrt{x} .\) (b) What value of \(a\) gives a critical point at \(x=5\) ? Does \(f(x)\) have a local maximum or a local minimum at this critical point?

Step-by-Step Solution

Answer

The value of \(a\) is \(2\sqrt{5}\), and the critical point at \(x=5\) is a local minimum.

Step 1: Find the derivative of the function

The function given is \( f(x) = x - a\sqrt{x} \). To find the critical points, we first need to find the derivative of the function. Using the power rule, the derivative of \( x \) is 1, and for \( a\sqrt{x} \) or \( ax^{1/2} \), the derivative is \( \frac{a}{2\sqrt{x}} \) or \( \frac{a}{2} x^{-1/2} \). Therefore, the derivative \( f'(x) \) is: \[ f'(x) = 1 - \frac{a}{2\sqrt{x}}. \]

Step 2: Solve for critical points

Critical points occur where the derivative \( f'(x) = 0 \) or is undefined. In our case, we solve the equation:\[ 1 - \frac{a}{2\sqrt{x}} = 0. \]Solving this equation for \( x \), we have:\[ 1 = \frac{a}{2\sqrt{x}}, \]\[ 2\sqrt{x} = a, \]\[ \sqrt{x} = \frac{a}{2}, \]\[ x = \left(\frac{a}{2}\right)^2. \]

Step 3: Solve for \(a\) when \(x = 5\)

Now, we need to find the value of \(a\) such that a critical point occurs at \( x = 5 \). Using our result from Step 2:\[ 5 = \left(\frac{a}{2}\right)^2 \]\[ \sqrt{5} = \frac{a}{2} \]\[ a = 2\sqrt{5}. \]

Step 4: Determine the nature of the critical point

To determine if the critical point is a local maximum or minimum, observe the second derivative. The second derivative \( f''(x) \) is necessary for this. First, differentiate \( f'(x) = 1 - \frac{a}{2\sqrt{x}} \),\[ f''(x) = \frac{a}{4x^{3/2}}. \]Since \( a > 0 \) and \( x > 0 \), \( f''(x) > 0 \), indicating that the function has a local minimum at this critical point.

Key Concepts

Derivative of a Function

The derivative of a function is a tool that helps us understand how the function is changing at any given point. Mathematically, it represents the rate of change or slope of the function at a point. When dealing with functions like \( f(x) = x - a\sqrt{x} \), the derivative tells us the behavior of \( f(x) \) concerning \( x \).
To find the derivative of our function, we apply rules such as the power rule. This rule is particularly useful with terms like \( x \) and \( a\sqrt{x} \).

• For a linear term \( x \), the derivative is quite straightforward: it's just 1.
• For terms involving powers, like \( a\sqrt{x} \) or \( ax^{1/2} \), the derivative is found using the formula \( \frac{a}{2} x^{-1/2} \), which simplifies to \( \frac{a}{2\sqrt{x}} \).

By combining these results, the derivative of the function \( f(x) = x - a\sqrt{x} \) becomes \( f'(x) = 1 - \frac{a}{2\sqrt{x}} \). This expression is crucial because it allows us to find critical points, which are locations on the graph where the function may achieve a maximum, a minimum, or change its direction.

Second Derivative Test

To determine the nature of critical points found using the first derivative, we employ the second derivative test. This test uses the second derivative of the function, \( f''(x) \), to identify whether a given critical point is a local minimum, a local maximum, or neither.
To execute the test, we calculate the second derivative of our original function. From the given function \( f'(x) = 1 - \frac{a}{2\sqrt{x}} \), the second derivative, \( f''(x) \), is determined by differentiating \( f'(x) \) once more.

• The resulting second derivative is \( f''(x) = \frac{a}{4x^{3/2}} \).
• This expression helps us see the concavity of the function around the critical point.

The second derivative test tells us:

• If \( f''(x) > 0 \), the function is concave upwards, indicating a local minimum at that point.
• If \( f''(x) < 0 \), the function is concave downwards, indicating a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive, and further analysis may be needed.

In the context of our problem, since \( a > 0 \) ensures \( f''(x) > 0 \) for all \( x > 0 \), this means the function has a local minimum at the found critical point.

Local Minimum and Maximum

When analyzing functions, local minima and maxima are points where the function reaches a peak or valley within a given interval. These points are critical for understanding the general shape and behavior of the function.
Local minima occur where the function drops to a low point in the interval before rising again. Local maxima happen where the function rises to a peak before dropping. The behavior of the function around its critical points often tells us their nature.

• At a local minimum, the function value is less than neighboring points.
• At a local maximum, the function value is greater than neighboring points.

Finding these points involves locating the critical points first (using the first derivative as we did) and then using the second derivative test to determine their nature.
For the function \( f(x) = x - a\sqrt{x} \) with critical points identified at \( x = \left(\frac{a}{2}\right)^2 \), the second derivative test informed us that these are indeed points of local minima.
This is because the test results in a positive second derivative \( f''(x) > 0 \), thus confirming the function is curving upwards, creating those low points. Understanding local minima and maxima assist greatly in graphing the function and in problems related to optimization, where finding the lowest or highest values is essential.