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Find the absolute maximum value and the absolute minimum value, if any, of each function. $$ f(x)=3 x^{2 / 3}-2 x \text { on }[0,3] $$

Short Answer

Expert verified
The absolute maximum value of the function \(f(x) = 3x^{\frac{2}{3}} - 2x\) on the interval \([0,3]\) is \(f(3) = 3\sqrt[3]{9} - 6\), and the absolute minimum value is \(f(0) = 0\).

Step by step solution

01

Identify the function and its derivative

We are given the function \(f(x) = 3x^{\frac{2}{3}} - 2x\) and need to find its derivative to identify potential candidates where extreme values might occur. Find the first derivative of the function, \(f'(x)\), using the power rule: \[\frac{d}{dx}(cx^n) = n(cx^{n-1})\]
02

Find the derivative of the function

Apply the power rule on \(f(x)\) to find its derivative. \[ f'(x) = \frac{d}{dx} \left( 3x^{\frac{2}{3}} - 2x \right) \] \[ f'(x) = \left(\frac{2}{3} \cdot 3x^{\frac{2}{3}-1}\right) - (2) \] \[ f'(x) = 2x^{\frac{-1}{3}} - 2 \]
03

Find the critical points

Critical points occur where the derivative is either 0 or undefined. In this case, set \(f'(x)\) equal to zero and solve for \(x\): \[ 2x^{\frac{-1}{3}} - 2 = 0 \] \[ x^{\frac{-1}{3}} = 1 \] \[ x = 1 \] So, \(x = 1\) is a critical point. There are no points in the given domain where \(f'(x)\) is undefined, so we don't have to worry about that.
04

Evaluate the function at critical points and endpoints

Now, we evaluate the function \(f(x)\) at the critical point \(x=1\) and endpoints \(x=0\) and \(x=3\): \[ f(0) = 3(0)^{\frac{2}{3}} - 2(0) = 0 \] \[ f(1) = 3(1)^{\frac{2}{3}} - 2(1) = 3 - 2 = 1 \] \[ f(3) = 3(3)^{\frac{2}{3}} - 2(3) = 3\sqrt[3]{9} - 6 \]
05

Determine absolute maximum and minimum values

Compare the function values from Step 4: Minimum value: The minimum value of the function is at \(x = 0\), where \(f(0) = 0\). Maximum value: The maximum value of the function is at \(x = 3\), where \(f(3) = 3\sqrt[3]{9} - 6\). Thus, the absolute maximum value of the function on the interval \([0,3]\) is \(f(3) = 3\sqrt[3]{9} - 6\) and the absolute minimum value is \(f(0) = 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
In calculus, critical points are key to identifying where a function's extreme values might occur. They are locations on the graph of a function where the derivative is either equal to zero or undefined.
This is because at these points, the function can potentially change direction, leading to local maxima or minima. However, critical points aren't the only places to look for extreme values.
Evaluating the function at the critical points gives us potential candidates for extreme values, which can be further classified as maximum, minimum, or sometimes even points of inflection, where the function's curvature changes.
Derivative
The derivative of a function is a fundamental concept in calculus. It represents how a function is changing at any given point and is often interpreted as the slope of the tangent line to the function at any point.
Mathematically, if you have a function \( f(x) \), its derivative \( f'(x) \) provides a mechanism to determine where the function will increase or decrease.
In this exercise, we used the derivative to find critical points by setting the derivative equal to zero, which helps to locate points of interest in the interval.
  • The power rule is a common method for finding derivatives, expressed as \( \frac{d}{dx}(cx^n) = n(cx^{n-1}) \).
  • The derivative can sometimes be undefined, especially for functions involving radicals or division by zero, and these points also need consideration as critical points.
Extreme Values
Extreme values, or extremum (singular), refer to the highest or lowest points in a specific interval of a function. These values are significant because they tell us about the limits of the output of a function in that space.
There are two types of extreme values: local (or relative) and absolute. Local extrema occur within a small neighborhood, while absolute extrema are the highest or lowest values over the entire interval.
Determining extreme values involves evaluating the function at critical points and endpoints of the interval. This step helps ensure that no potential extreme value is missed. For the given function, extreme values were sought by checking the function at critical points and endpoints within the specified interval.
Absolute Maximum and Minimum
In calculus, determining the absolute maximum and minimum values of a function on a particular interval is essential for understanding the function's behavior within that domain.
The absolute maximum value is the highest value a function reaches, and the absolute minimum is the lowest value on the interval.
To find these, we evaluate the function at all critical points and endpoints, comparing these values to identify the absolute extremes.
  • The process typically involves using critical points, derived from setting the derivative to zero, and comparing these with the function values at the interval's boundaries.
  • The given exercise found the absolute maximum value at \( x = 3 \) and the absolute minimum at \( x = 0 \), calculated by evaluating and comparing the function at these points.
Understanding absolute extrema is crucial for comprehensive analysis, particularly in optimization problems.

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