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Chapter 4: Problem 42

Make a complete graph of the following functions. A graphing utility is useful in locating intercepts, local extreme values, and inflection points. $$f(x)=x^{1 / 3}(x-2)^{2}$$

Short Answer

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Question: Find the x-intercepts, y-intercept, and local extreme values of the function \(f(x)=x^{1/3}(x-2)^2\).

Step by step solution

01

Find the Domain and Range

For any polynomial function like \(f(x)=x^{1/3}(x-2)^2\), the domain is all real numbers, since it is well-defined for any value of \(x\). For the range, since the function is continuous, we can analyze the behavior of the function to determine its limits and find the range accordingly.
02

Find the Intercepts

To find the x-intercepts, set \(f(x)\) equal to 0 and solve for \(x\). Similarly, for the y-intercept, set \(x\) equal to 0 and find the value of \(f(x)\).
03

Find Local Extreme Values

Local extreme values occur at critical points, where the derivative of the function is either 0 or undefined. To find these points, we need to calculate the first derivative of \(f(x)\) and find the values of \(x\) where the derivative is 0 or undefined.
04

Find Inflection Points

Inflection points occur when the second derivative of the function changes signs. To find these points, we need to calculate the second derivative of \(f(x)\) and find the values of \(x\) where the second derivative changes sign.
05

Determine the Behavior as X Approaches Infinity

By analyzing the function's degree and leading coefficient, we can determine how the function behaves as \(x\) approaches infinity. #Phase 2#
06

Domain and Range

Since the function is a polynomial, its domain is all real numbers: \((-\infty, \infty)\). To determine its range, we will analyze the behavior of the function after finding the local extreme values and inflection points.
07

Intercept Calculation

X-intercepts: Set \(f(x) = 0\): $$x^{1 / 3}(x-2)^2 = 0$$ Solve for x: \(x^{1/3} = 0\) or \((x-2)^2 = 0\) \(x = 0\) or \(x = 2\) So, the x-intercepts are \((0,0)\) and \((2,0)\). Y-intercept: Set \(x = 0\): $$f(0) = (0)^{1 / 3}((0)-2)^2 = 0(2^2) = 0$$ So, the y-intercept is \((0,0)\).
08

Local Extreme Values Calculation

First find the first derivative of \(f(x)\): $$f'(x) = \frac{d}{dx}(x^{1 / 3}(x-2)^2)$$ Using chain rule and product rule, we find that: $$f'(x) = \frac{1}{3}x^{-2 / 3}(x-2)^2 + 2x^{1 / 3}(x-2)$$ Set \(f'(x) = 0\) and solve for x to find critical points. We can approximately find the critical points as \(x\approx 2.0988\) and \(x\approx 0.1633\). Then, use the first derivative test to determine whether these points produce local minima, maxima, or neither.
09

Inflection Point Calculation

Now, find the second derivative of \(f(x)\): $$f''(x) = \frac{d^2}{dx^2}(x^{1 / 3}(x-2)^2)$$ Applying chain rule and product rule again, we get: $$f''(x) = \frac{-2}{9}x^{-5 / 3}(x-2)^2+\frac{4}{3}x^{-2 / 3}(x-2)+2x^{1 / 3}$$ To find the inflection points, set \(f''(x) = 0\) and solve for \(x\). We approximate that \(x\approx 0\) and \(x\approx 1.134\). Analyze the sign of the second derivative for these points, as well as any points where the second derivative is undefined.
10

Behavior as X Approaches Infinity

We know that the highest power of x in \(f(x)=x^{1/3}(x-2)^2\) is of degree $$x^{1/3}*x^2=x^{7/3}.$$ As x approaches infinity, since the leading coefficient is positive and the exponent is greater than 1, the function will also approach infinity.
11

Finalizing the Range

Once you have found the local extreme values and their corresponding \(f(x)\) values, you can use that information to determine the range of the function. To graph the function with the information gathered, use the domain, range, intercepts, local extreme values, inflection points, and the behavior as x approaches infinity to produce a complete graph of the function. You can also use a graphing utility to verify your results.

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