Absolute Maxima And Minima Using The First Derivative Test

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Understanding the First Derivative Test

The first derivative test is a fundamental tool in calculus for identifying local maxima and minima of a function. This powerful method leverages the relationship between a function's derivative and its increasing or decreasing behavior. By analyzing the sign changes of the first derivative, we can pinpoint critical points where the function potentially reaches its peak or valley. These critical points, where the derivative is either zero or undefined, are the prime candidates for local extrema. However, to determine the absolute maxima and minima over a closed interval, we need to extend our analysis beyond the critical points. Absolute extrema represent the highest and lowest function values within the specified interval, and they may occur at either critical points or the endpoints of the interval.

The first derivative test plays a crucial role in optimization problems, where the goal is to find the maximum or minimum value of a function subject to certain constraints. In various fields, such as engineering, economics, and physics, optimization problems arise frequently. For example, an engineer might seek to design a bridge that minimizes material usage while maintaining structural integrity, or an economist might aim to maximize profit given production costs and market demand. By employing the first derivative test and related techniques, we can systematically solve these optimization problems and make informed decisions. When seeking to find absolute maxima and minima, remember that these extreme values represent the highest and lowest points of a function within a given domain, which can have significant practical implications in real-world applications.

The process of finding absolute extrema using the first derivative test involves several key steps. First, we must determine the critical points of the function within the given interval. These critical points are where the first derivative is either zero or undefined. Setting the derivative equal to zero helps us find stationary points, while points where the derivative is undefined may indicate sharp corners or vertical tangents. Next, we evaluate the function at each critical point and at the endpoints of the interval. The largest of these values is the absolute maximum, and the smallest is the absolute minimum. This comprehensive approach ensures that we capture the global extreme values within the specified domain. For instance, consider a roller coaster track modeled by a function; finding the absolute maximum and minimum points would correspond to the highest and lowest elevations of the track, which are crucial for safety and design considerations. Therefore, the first derivative test is an indispensable tool for locating these critical values and optimizing various real-world scenarios.

Applying the First Derivative Test to f(x)=x2ext1−x2f(x) = x^2 ext{\sqrt{1-x^2}} on [−1,1][-1, 1]

To find the absolute maxima and minima of the function f(x)=x2ext1−x2f(x) = x^2 ext{\sqrt{1-x^2}} on the closed interval [−1,1][-1, 1], we will meticulously apply the first derivative test. This process involves several key steps, starting with finding the derivative of the function, identifying critical points, and finally evaluating the function at these points and the interval endpoints.

Our first step is to compute the derivative of f(x)f(x). Using the product rule and the chain rule, we have:

f'(x) = rac{d}{dx} [x^2 ext{\sqrt{1-x^2}}]

= 2x ext{\sqrt{1-x^2}} + x^2 rac{d}{dx} [ ext{\sqrt{1-x^2}}]

= 2x ext{\sqrt{1-x^2}} + x^2 rac{1}{2 ext{\sqrt{1-x^2}}} (-2x)

= 2x ext{\sqrt{1-x^2}} - rac{x^3}{ ext{\sqrt{1-x^2}}}

To simplify this expression, we find a common denominator:

f'(x) = rac{2x(1-x^2) - x^3}{ ext{\sqrt{1-x^2}}}

= rac{2x - 2x^3 - x^3}{ ext{\sqrt{1-x^2}}}

= rac{2x - 3x^3}{ ext{\sqrt{1-x^2}}}

Now we can factor out an xx from the numerator:

f'(x) = rac{x(2 - 3x^2)}{ ext{\sqrt{1-x^2}}}

The next crucial step is to find the critical points of the function. These are the points where the derivative is either zero or undefined. Setting the derivative equal to zero, we get:

rac{x(2 - 3x^2)}{ ext{\sqrt{1-x^2}}} = 0

This equation is satisfied when the numerator is zero, i.e.,

x(2−3x2)=0x(2 - 3x^2) = 0

This gives us three possible values for xx:

  1. x=0x = 0
  2. 2 - 3x^2 = 0 ightarrow 3x^2 = 2 ightarrow x^2 = rac{2}{3} ightarrow x = ext{\pm\sqrt{\frac{2}{3}}}

So the critical points where the derivative is zero are x=0x = 0, x=ext23x = ext{\sqrt{\frac{2}{3}}}, and x=−ext23x = - ext{\sqrt{\frac{2}{3}}}.

Next, we need to consider where the derivative is undefined. This occurs when the denominator is zero:

$ ext{\sqrt{1-x^2}} = 0$

1−x2=01 - x^2 = 0

x2=1x^2 = 1

x=ext±1x = ext{\pm1}

Thus, the derivative is undefined at x=−1x = -1 and x=1x = 1. These are also the endpoints of our interval [−1,1][-1, 1].

Now we have identified all the critical points within the interval [−1,1][-1, 1]: x=−1x = -1, x=−ext23x = - ext{\sqrt{\frac{2}{3}}}, x=0x = 0, x=ext23x = ext{\sqrt{\frac{2}{3}}}, and x=1x = 1. The final step is to evaluate the original function f(x)f(x) at these critical points to determine the absolute maxima and minima. This comprehensive approach ensures we capture the function's extreme values within the specified domain.

Evaluating the Function at Critical Points and Endpoints

After identifying the critical points and endpoints within the interval [−1,1][-1, 1], the next step in finding the absolute maxima and minima of f(x)=x2ext1−x2f(x) = x^2 ext{\sqrt{1-x^2}} is to evaluate the function at these key points. This will help us determine the maximum and minimum values of the function within the given interval. The critical points we found are x=−1x = -1, x=−ext23x = - ext{\sqrt{\frac{2}{3}}}, x=0x = 0, x=ext23x = ext{\sqrt{\frac{2}{3}}}, and x=1x = 1.

  1. Evaluate f(x)f(x) at x=−1x = -1:

    f(−1)=(−1)2ext1−(−1)2=1ext1−1=1imes0=0f(-1) = (-1)^2 ext{\sqrt{1 - (-1)^2}} = 1 ext{\sqrt{1 - 1}} = 1 imes 0 = 0

  2. Evaluate f(x)f(x) at x=−ext23x = - ext{\sqrt{\frac{2}{3}}}:

    f(- ext{\sqrt{\frac{2}{3}}}) = (- ext{\sqrt{\frac{2}{3}}})^2 ext{\sqrt{1 - (- ext{\sqrt{\frac{2}{3}}})^2}} = rac{2}{3} ext{\sqrt{1 - rac{2}{3}}} = rac{2}{3} ext{\sqrt{\frac{1}{3}}} = rac{2}{3} rac{1}{ ext{\sqrt{3}}} = rac{2}{3 ext{\sqrt{3}}} = rac{2 ext{\sqrt{3}}}{9}

  3. Evaluate f(x)f(x) at x=0x = 0:

    f(0)=(0)2ext1−(0)2=0ext1=0f(0) = (0)^2 ext{\sqrt{1 - (0)^2}} = 0 ext{\sqrt{1}} = 0

  4. Evaluate f(x)f(x) at x=ext23x = ext{\sqrt{\frac{2}{3}}}:

    f( ext{\sqrt{\frac{2}{3}}}) = ( ext{\sqrt{\frac{2}{3}}})^2 ext{\sqrt{1 - ( ext{\sqrt{\frac{2}{3}}})^2}} = rac{2}{3} ext{\sqrt{1 - rac{2}{3}}} = rac{2}{3} ext{\sqrt{\frac{1}{3}}} = rac{2}{3} rac{1}{ ext{\sqrt{3}}} = rac{2}{3 ext{\sqrt{3}}} = rac{2 ext{\sqrt{3}}}{9}

  5. Evaluate f(x)f(x) at x=1x = 1:

    f(1)=(1)2ext1−(1)2=1ext1−1=1imes0=0f(1) = (1)^2 ext{\sqrt{1 - (1)^2}} = 1 ext{\sqrt{1 - 1}} = 1 imes 0 = 0

By evaluating the function at these points, we have:

  • f(−1)=0f(-1) = 0
  • f(- ext{\sqrt{\frac{2}{3}}}) = rac{2 ext{\sqrt{3}}}{9} ext{ \approx 0.3849}
  • f(0)=0f(0) = 0
  • f( ext{\sqrt{\frac{2}{3}}}) = rac{2 ext{\sqrt{3}}}{9} ext{ \approx 0.3849}
  • f(1)=0f(1) = 0

Comparing these values, we can identify the absolute maximum and absolute minimum values of the function within the interval [−1,1][-1, 1]. This systematic approach ensures that we accurately determine the extreme values, which is a critical aspect of optimization problems and calculus in general.

Determining Absolute Maxima and Minima

Having evaluated the function f(x)=x2ext1−x2f(x) = x^2 ext{\sqrt{1-x^2}} at all critical points and endpoints within the interval [−1,1][-1, 1], we can now definitively determine the absolute maxima and absolute minima. By comparing the function values at these points, we identify the highest and lowest values attained by the function within the specified interval. Our calculations yielded the following function values:

  • f(−1)=0f(-1) = 0
  • f(- ext{\sqrt{\frac{2}{3}}}) = rac{2 ext{\sqrt{3}}}{9} ext{ \approx 0.3849}
  • f(0)=0f(0) = 0
  • f( ext{\sqrt{\frac{2}{3}}}) = rac{2 ext{\sqrt{3}}}{9} ext{ \approx 0.3849}
  • f(1)=0f(1) = 0

From these values, it is clear that the function attains its maximum value at x=−ext23x = - ext{\sqrt{\frac{2}{3}}} and x=ext23x = ext{\sqrt{\frac{2}{3}}}, where f(x) = rac{2 ext{\sqrt{3}}}{9}. Therefore, the absolute maximum value of the function on the interval [−1,1][-1, 1] is rac{2 ext{\sqrt{3}}}{9}. This represents the highest point the function reaches within the given domain.

On the other hand, the function attains its minimum value at x=−1x = -1, x=0x = 0, and x=1x = 1, where f(x)=0f(x) = 0. Thus, the absolute minimum value of the function on the interval [−1,1][-1, 1] is 00. This represents the lowest point the function reaches within the given domain. The identification of absolute extrema is a crucial outcome of the first derivative test, enabling us to understand the global behavior of the function within the specified interval.

In summary, by systematically applying the first derivative test, we have successfully identified the absolute maxima and minima of the function f(x)=x2ext1−x2f(x) = x^2 ext{\sqrt{1-x^2}} on the interval [−1,1][-1, 1]. The absolute maximum value is rac{2 ext{\sqrt{3}}}{9}, occurring at x=−ext23x = - ext{\sqrt{\frac{2}{3}}} and x=ext23x = ext{\sqrt{\frac{2}{3}}}, and the absolute minimum value is 00, occurring at x=−1x = -1, x=0x = 0, and x=1x = 1. This comprehensive analysis demonstrates the power and utility of calculus techniques in determining the extreme values of functions, which has broad applications in various fields.