Problem 36 - Calculus Single Variable

Question

Find each critical point \(c\) of the given function \(f\). Then use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=3 x^{1 / 3}-5 x^{1 / 5} $$

Step-by-Step Solution

Answer

The critical point is \( x = 0 \), but it's neither a local max nor min.

Step 1: Find the Derivative

First, calculate the derivative of the function \( f(x) = 3x^{1/3} - 5x^{1/5} \). To do this, apply the power rule for derivatives. Since the derivative of \( x^n \) is \( nx^{n-1} \), we get the derivative of the function as follows:\[ \frac{d}{dx}(3x^{1/3}) = 1x^{-2/3}, \]\[ \frac{d}{dx}(-5x^{1/5}) = -x^{-4/5}. \]Thus, \[ f'(x) = 1x^{-2/3} - x^{-4/5}. \]

Step 2: Set the Derivative to Zero

To find the critical points, set the derivative equal to zero: \[ 1x^{-2/3} - x^{-4/5} = 0. \]This becomes \[ x^{-2/3} = x^{-4/5}. \]

Step 3: Solve the Equation

Rewriting the equation from Step 2 gives:\[ x^{4/5} = x^{2/3}. \]Convert both expressions to a common power or base. Here, you can solve by rewriting:\[ (x^{4/5})^{15} = (x^{2/3})^{18} \]. Thus, \[ x^{12} = x^{20}. \]This implies either \( x = 0 \) or the simplification yields that \( x^{8} = 1 \). Therefore, critical point is \( x = 0 \), as other values from roots of unity do not differ by an influence greater than at endpoint.

Step 4: Apply the First Derivative Test

To determine what type of critical point \( x = 0 \) is, consider intervals around \( 0 \). - Choose test points, for example, \( x = -1 \) and \( x = 1 \) to test the sign of \( f'(x) \) in these regions. - For \( x = -1 \), \( f'(-1) = -1 - 1 = -2 \), which is negative. - For \( x = 1 \), \( f'(1) = 1 - 1 = 0 \), but check \( x=2 \) or \( x=0.5 \) for clarity - If \( f'(x) \) changes from negative to positive, the function has a local minimum at this point.Since \( f'(x) \) does not sign switch, we can shorthand confirm concavity absence conclusion to assume 'concave alignment or multifactor'. Therefore, no definitive local min or max inferred.

Key Concepts

First Derivative Test

The First Derivative Test is a vital tool in calculus used to analyze the behavior of a function around its critical points. A critical point occurs where the derivative of the function equals zero or where the function is not differentiable. By evaluating the derivative before and after these points, we understand the nature of these critical points.

• If the derivative changes from positive to negative around the critical point, the function has a local maximum at the point.
• If the derivative changes from negative to positive, the function has a local minimum.
• If the derivative does not change signs, then the critical point is neither a maximum nor a minimum, also known as an inflection point.

To apply this test, carefully choose test points in intervals around the critical point and evaluate the sign of the derivative. This provides insight into the function's growth or shrinkage behavior.

Power Rule for Derivatives

The power rule for derivatives is one of the fundamental tools in calculus for finding the derivative of a function. If you have a function of the form \[ f(x) = x^n \] its derivative is calculated as \[ f'(x) = nx^{n-1}. \]This rule simplifies the process of differentiation and is applicable as long as the exponent \( n \) is a real number.
For example, applying the power rule to the function \( f(x) = 3x^{1/3} - 5x^{1/5} \), results in:

• For \( 3x^{1/3} \), the power rule results in the derivative \( 1x^{-2/3}.\)
• For \( -5x^{1/5} \), you derive \( -1x^{-4/5}.\)

Thus, the derivative \( f'(x) = 1x^{-2/3} - x^{-4/5} \) can be used to find critical points and analyze the behavior of the original function.

Local Maximum and Minimum

In calculus, local maximums and minimums are essential concepts that help understand the peaks and troughs of a function. A local maximum occurs when a function's value at a certain point is higher than any nearby points. Conversely, a local minimum occurs when the function's value is lower than any neighboring points.
For determining these, we primarily use critical points obtained from the function's derivative.

• A local maximum is recognized when the function's derivative transitions from positive to negative at that point.
• A local minimum is identified when the derivative changes from negative to positive.
• If there's no sign change, the point can be an inflection point rather than a maximum or minimum.

These insights help in sketching the graph of functions and understanding their real-world implications. In our exercise, even after using the First Derivative Test on the critical point \( x = 0 \), the function did not reveal a definitive local maximum or minimum, showcasing a scenario where signs and predictions via derivatives are critical.