Find The Absolute Maximum Of $f(x)=\sqrt{4-x^2}$ On $[-2,2]$.A. \[$-2\$\]B. \[$0\$\]C. \[$1\$\]D. \[$2\$\]

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Introduction

In calculus, finding the absolute maximum of a function on a closed interval is a crucial problem. It involves determining the maximum value of the function within the given interval. In this article, we will focus on finding the absolute maximum of the function $f(x)=\sqrt{4-x^2}$ on the interval $[-2,2]$. We will use various techniques to solve this problem and provide a step-by-step solution.

Understanding the Function

The given function is $f(x)=\sqrt{4-x^2}$. This is a square root function, and its domain is restricted to the interval $[-2,2]$ due to the presence of the square root. The function is defined as long as the expression inside the square root is non-negative.

Analyzing the Function

To find the absolute maximum of the function, we need to analyze its behavior on the given interval. We can start by finding the critical points of the function, which are the points where the derivative of the function is zero or undefined.

Finding the Derivative

To find the derivative of the function, we can use the chain rule. The derivative of the square root function is given by:

ddxu=12ududx\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\frac{du}{dx}

In this case, $u = 4-x^2$. Therefore, the derivative of the function is:

f′(x)=124−x2ddx(4−x2)f'(x) = \frac{1}{2\sqrt{4-x^2}}\frac{d}{dx}(4-x^2)

f′(x)=124−x2(−2x)f'(x) = \frac{1}{2\sqrt{4-x^2}}(-2x)

Finding the Critical Points

The critical points of the function are the points where the derivative is zero or undefined. In this case, the derivative is undefined when $4-x^2 = 0$, which gives $x = \pm 2$. The derivative is also zero when $-2x = 0$, which gives $x = 0$.

Evaluating the Function at the Critical Points

To find the absolute maximum of the function, we need to evaluate the function at the critical points and at the endpoints of the interval.

  • At $x = -2$, the function is $f(-2) = \sqrt{4-(-2)^2} = \sqrt{0} = 0$.
  • At $x = 0$, the function is $f(0) = \sqrt{4-0^2} = \sqrt{4} = 2$.
  • At $x = 2$, the function is $f(2) = \sqrt{4-2^2} = \sqrt{0} = 0$.

Comparing the Values

To find the absolute maximum of the function, we need to compare the values of the function at the critical points and at the endpoints of the interval.

  • The value of the function at $x = -2$ is $0$.
  • The value of the function at $x = 0$ is $2$.
  • The value of the function at $x = 2$ is $0$.

Conclusion

Based on the analysis, the absolute maximum of the function $f(x)=\sqrt{4-x^2}$ on the interval $[-2,2]$ is $2$, which occurs at $x = 0$.

Answer

The correct answer is:

  • B. ${0\$}

Discussion

This problem involves finding the absolute maximum of a function on a closed interval. We used various techniques, including finding the derivative, finding the critical points, and evaluating the function at the critical points and at the endpoints of the interval. The absolute maximum of the function occurs at the point where the function has the largest value.

Related Problems

  • Finding the absolute minimum of a function on a closed interval.
  • Finding the local maximum and minimum of a function.
  • Using the second derivative test to determine the nature of the critical points.

References

  • Calculus by Michael Spivak.
  • Calculus by James Stewart.
  • Calculus by David Guichard.

Keywords

  • Absolute maximum
  • Closed interval
  • Critical points
  • Derivative
  • Square root function
    Q&A: Finding the Absolute Maximum of a Function on a Closed Interval ====================================================================

Introduction

In our previous article, we discussed how to find the absolute maximum of a function on a closed interval. In this article, we will provide a Q&A section to help clarify any doubts and provide additional information on this topic.

Q: What is the absolute maximum of a function?

A: The absolute maximum of a function is the largest value that the function attains on a given interval.

Q: How do I find the absolute maximum of a function on a closed interval?

A: To find the absolute maximum of a function on a closed interval, you need to follow these steps:

  1. Find the critical points of the function by setting the derivative equal to zero and solving for x.
  2. Evaluate the function at the critical points and at the endpoints of the interval.
  3. Compare the values of the function at the critical points and at the endpoints of the interval to determine the absolute maximum.

Q: What are critical points?

A: Critical points are the points where the derivative of the function is zero or undefined. These points are important because they can be local maxima or minima, or they can be points where the function changes from increasing to decreasing or vice versa.

Q: How do I find the derivative of a function?

A: To find the derivative of a function, you can use the power rule, the product rule, the quotient rule, and the chain rule. The power rule states that if f(x) = x^n, then f'(x) = nx^(n-1). The product rule states that if f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x). The quotient rule states that if f(x) = u(x)/v(x), then f'(x) = (u'(x)v(x) - u(x)v'(x)) / v(x)^2. The chain rule states that if f(x) = g(h(x)), then f'(x) = g'(h(x)) * h'(x).

Q: What is the difference between a local maximum and an absolute maximum?

A: A local maximum is the largest value of a function on a given interval, but it may not be the largest value on the entire domain of the function. An absolute maximum, on the other hand, is the largest value of a function on its entire domain.

Q: Can a function have more than one absolute maximum?

A: No, a function can only have one absolute maximum on a given interval.

Q: How do I determine if a critical point is a local maximum or a local minimum?

A: To determine if a critical point is a local maximum or a local minimum, you can use the second derivative test. If the second derivative is positive at the critical point, then it is a local minimum. If the second derivative is negative at the critical point, then it is a local maximum.

Q: What is the second derivative test?

A: The second derivative test is a method used to determine the nature of a critical point. If the second derivative is positive at the critical point, then it is a local minimum. If the second derivative is negative at the critical point, then it is a local maximum.

Q: Can I use the second derivative test on a function that is not differentiable at a point?

A: No, you cannot use the second derivative test on a function that is not differentiable at a point.

Q: What are some common mistakes to avoid when finding the absolute maximum of a function?

A: Some common mistakes to avoid when finding the absolute maximum of a function include:

  • Not evaluating the function at the endpoints of the interval.
  • Not comparing the values of the function at the critical points and at the endpoints of the interval.
  • Not using the second derivative test to determine the nature of the critical points.

Conclusion

Finding the absolute maximum of a function on a closed interval is an important problem in calculus. By following the steps outlined in this article, you can find the absolute maximum of a function on a closed interval. Remember to evaluate the function at the critical points and at the endpoints of the interval, and to compare the values of the function at the critical points and at the endpoints of the interval to determine the absolute maximum.

Related Problems

  • Finding the absolute minimum of a function on a closed interval.
  • Finding the local maximum and minimum of a function.
  • Using the second derivative test to determine the nature of the critical points.

References

  • Calculus by Michael Spivak.
  • Calculus by James Stewart.
  • Calculus by David Guichard.

Keywords

  • Absolute maximum
  • Closed interval
  • Critical points
  • Derivative
  • Second derivative test