Problem 38 - Calculus Early Transcendentals

Question

a. Find the critical points of \(f\) on the given interval. b. Determine the absolute extreme values of \(f\) on the given interval when they exist. c. Use a graphing utility to confirm your conclusions. $$f(x)=(x+1)^{4 / 3} \text { on }[-9,7]$$

Step-by-Step Solution

Answer

Answer: The absolute minimum value of the function is \(0\) at \(x=-1\), and the absolute maximum values are \(16\) at \(x=-9\) and \(x=7\).

Step 1: Find the derivative of the function

To find the critical points of the function, we first need to find the derivative of the given function \(f(x)=(x+1)^{\frac{4}{3}}\). We can use the power rule, which states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). Applying the power rule to the given function, we get: $$f'(x)=\frac{4}{3}(x+1)^{\frac{4}{3}-1} = \frac{4}{3}(x+1)^{\frac{1}{3}}$$

Step 2: Find the critical points

Critical points are found where the derivative \(f'(x)\) is either equal to zero or is undefined. In this case, the derivative will never be equal to zero, but it will be undefined when the exponent of \((x+1)\) is negative for any value of \(x\). Let's solve for x to find out if it happens in the given interval \([-9, 7]\): $$ (x+1)^{\frac{1}{3}}=0$$ We see that the above equation cannot be satisfied, so the critical point must occur when the exponent of \((x+1)\) is undefined. The exponent will be undefined when the base, \((x+1)\), is negative, and the exponent is a fraction. The base \((x+1)\) will be negative when \(x<-1\). Now we can check if this interval intersects with the given interval \([-9,7]\). Since \((-1)\) lies in the interval \([-9,7]\), then we can conclude that there is one critical point at \(x=-1\).

Step 3: Determine the absolute extreme values

To determine the absolute extreme values of the function \(f(x)\) on the interval \([-9,7]\), we need to evaluate the original function at the critical point \(x=-1\), as well as at the endpoints of the interval. Then, we will compare these values to determine the absolute maximum and minimum. Let's compute the values of the function at these three points: $$f(-9)=(-9+1)^{\frac{4}{3}}=(-8)^{\frac{4}{3}}=16$$ $$f(-1)=(-1+1)^{\frac{4}{3}}=(0)^{\frac{4}{3}}=0$$ $$f(7)=(7+1)^{\frac{4}{3}}=(8)^{\frac{4}{3}}=16$$ From these values, we can conclude that there is an absolute minimum value of \(0\) at \(x=-1\), and the absolute maximum values of \(16\) at \(x=-9\) and \(x=7\).

Step 4: Use a graphing utility to confirm

Plot the function \(f(x)=(x+1)^{\frac{4}{3}}\) on a graphing utility and observe the behavior of the function on the interval \([-9,7]\). You will see that the function has an absolute minimum value of \(0\) at \(x=-1\), and the absolute maximum values of \(16\) at \(x=-9\) and \(x=7\), thus confirming our conclusions.

Key Concepts

What are Critical Points?

Critical points are important for understanding the behavior of a function. They are where the derivative of the function is either zero or undefined. These points typically correspond to local maxima, minima, or points of inflection. To find critical points, follow these steps:

• First, determine the derivative of the function.
• Set the derivative equal to zero and solve for the variable.
• Identify where the derivative is undefined by checking where the derivative involves any division by zero or other undefined expressions.

In the example function \(f(x) = (x+1)^{\frac{4}{3}}\), the critical points occur when the derivative \(f'(x) = \frac{4}{3}(x+1)^{\frac{1}{3}}\) is undefined. Here, the point \(x = -1\) is critical because the base \((x+1)\) is zero, leading to the derivative being undefined at this point.

Understanding Absolute Extreme Values

Absolute extreme values refer to the highest and lowest points that a function can reach over a given interval. They are called the absolute maximum and minimum. To find these values, evaluate the function at:

• The critical points you have found.
• The endpoints of the interval.

This ensures you consider all possible locations for extremes on the interval. In our function \(f(x) = (x+1)^{\frac{4}{3}}\) on the interval \([-9, 7]\), we evaluate the function at critical point \(x = -1\) and the endpoints \(x = -9\) and \(x = 7\). Calculating gives \(f(-9) = 16\), \(f(-1) = 0\), and \(f(7) = 16\). Thus, the absolute minimum value is 0 at \(x = -1\) and the absolute maximum value is 16 at both \(x = -9\) and \(x = 7\).

The Derivative and Its Role

The derivative of a function, denoted \(f'(x)\), measures the rate at which the function is changing at any given point. It reflects the slope of the tangent line to the function at any point. The derivative is crucial for:

• Finding critical points.
• Determining the behavior and shape of the function curve.
• Solving optimization problems.

For the function \(f(x) = (x+1)^{\frac{4}{3}}\), its derivative \(f'(x) = \frac{4}{3}(x+1)^{\frac{1}{3}}\) was used to understand where the function's slope changes. By examining where the derivative is zero or undefined, we gain insights into the critical points and, consequently, the extreme values the function can reach. Therefore, mastering the concept of derivatives is essential in calculus for analyzing and optimizing functions.