Problem 20 - Essential Calculus

Question

Locate the absolute extrema of the function on the closed interval. $$ g(x)=\sqrt[3]{x},[-1,1] $$

Step-by-Step Solution

Answer

The absolute maximum of the function \(g(x) = \sqrt[3]{x}\) on the interval [-1,1] is at \(x = 1\) and the absolute minimum is at \(x = -1\).

Step 1: Find the derivative of the function

Apply the power rule for differentiation (\(d/dx[x^n] = n x^{n-1}\)) to find the derivative of \(g(x)\). Here, \(n=1/3\), so the derivative \(g'(x) = (1/3)x^{-2/3}\).

Step 2: Find critical points

To find critical points, set the derivative equal to zero and solve for \(x\). However, for this function, \(g'(x)\) is undefined at \(x=0\), making \(x=0\) a critical point.

Step 3: Evaluate the function at critical points and the end points

The extrema of the function on the interval will be at the critical points or the end points of the interval. So, evaluate the function at these points: \(g(-1)\), \(g(0)\), and \(g(1)\). Since \(g(x)=\sqrt[3]{x}\), \(g(-1)=-1\), \(g(0)=0\), and \(g(1)=1\).

Step 4: Identify the absolute extrema

The values of g at the points from step 3 are respective to those points. The absolute maximum of g on the interval is at x = 1 and the absolute minimum is at x = -1.

Key Concepts

Critical Points

Critical points in calculus are where the derivative of a function is either zero or undefined. These points are considered because they can indicate where local maxima, local minima, or points of inflection occur. In this exercise, we calculated the derivative of the function \( g(x) = \sqrt[3]{x} \), which is \( g'(x) = \frac{1}{3}x^{-2/3} \). Here, the critical point is found by noting that \( g'(x) \) is undefined at \( x = 0 \). Since the derivative does not exist at this point, \( x = 0 \) is a critical point for the function \( g(x) \). It’s important to check these points along with endpoints to identify extrema on a closed interval.

Derivative

A derivative represents the rate of change of a function relative to one of its variables. It provides critical information about the behavior of a function. In simpler terms, it's like making predictions on how a function value changes as \( x \) changes. For the function \( g(x) = \sqrt[3]{x} \), the derivative using the power rule is \( g'(x) = \frac{1}{3}x^{-2/3} \). The derivative helps us locate critical points where the function might achieve extrema. In scenarios where the derivative equals zero or is undefined, significant changes in the function's behavior can occur.

Closed Interval

A closed interval includes all the numbers between and including its endpoints. It is denoted in brackets like \([-1, 1]\), incorporating both the start and end values. The closed interval is crucial when finding absolute extrema because it dictates that we evaluate the function not only at critical points but also at these interval endpoints. Calculating the values at \(x = -1\), \(x = 0\), and \(x = 1\) ensures we consider every potential point on the interval that can be an absolute maximum or minimum.

Power Rule

The Power Rule is a fundamental technique in differential calculus used to differentiate functions of the form \(x^n\). According to this rule, the derivative of \(x^n\) is \(nx^{n-1}\). Applying this principle greatly simplifies the process of differentiation. In our exercise, we used the power rule to find the derivative of \(g(x) = \sqrt[3]{x}\), which can be rewritten as \(x^{1/3}\). So using the power rule, \(g'(x) = \frac{1}{3}x^{-2/3}\). This rule is invaluable for differentiating a wide range of polynomial functions quickly and accurately.