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Chapter 8: Problem 5

Show that the absolute minimum of \(f(x)=\sqrt{25-x^{2}}\) on [-5,5] is 0 and the absolute maximum is 5.

Short Answer

Expert verified
Answer: The absolute minimum value is 0 and the absolute maximum value is 5.

Step by step solution

01

Find the first derivative of f(x)

First, we differentiate the function with respect to x. Since \(f(x) = \sqrt{25 - x^2}\), we can rewrite this as \(f(x) = (25 - x^2)^{1/2}\). Then, applying the chain rule to find the derivative: $$f'(x) = \frac{1}{2}(25 - x^2)^{-1/2}(-2x) = \frac{-x}{\sqrt{25 - x^2}}$$
02

Determine the critical points

Now we must find the critical points of the function, so we must find where \(f'(x) = 0\) or \(f'(x)\) is undefined. The first derivative is undefined when the denominator is zero, which happens when x = 卤5. So, let's find where the first derivative equals zero: $$\frac{-x}{\sqrt{25 - x^2}} = 0$$ Multiplying both sides of the equation by \(\sqrt{25 - x^2}\), we get: $$-x = 0$$ So, we have one critical point at \(x=0\).
03

Evaluate the function at critical points and endpoints

We will now evaluate the function \(f(x)\) at all critical points and the two endpoints (-5 and 5) since the function is continuous everywhere in \([-5, 5]\). The two critical points are x = -5, 0, and 5. $$f(-5) = \sqrt{25 - (-5)^2} = \sqrt{0} = 0$$ $$f(0) = \sqrt{25 - 0^2} = \sqrt{25} = 5$$ $$f(5) = \sqrt{25 - 5^2} = \sqrt{0} = 0$$
04

Determine the absolute minimum and maximum

Comparing the values of the function, we see that: - Absolute minimum is 0, which occurs at x = -5 and x = 5. - Absolute maximum is 5, which occurs at x = 0. Hence, the absolute minimum of \(f(x)\) on the interval \([-5, 5]\) is 0, and the absolute maximum is 5.

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Key Concepts

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

Critical Points
Critical points are specific values in the domain of a function where its derivative is either zero or undefined. These points are essential for finding the local maxima and minima, as well as identifying potential locations of absolute extrema on a given interval.
  • To find critical points, follow these steps: Calculate the derivative of the function.
  • Solve the equation where the derivative is zero.
  • Identify points where the derivative is undefined.
For the function \(f(x) = \sqrt{25 - x^2}\), the derivative is \(f'(x) = \frac{-x}{\sqrt{25 - x^2}}\). Solving \(f'(x) = 0\) gives us a critical point at \(x = 0\). Since the expression is undefined when the denominator \(\sqrt{25 - x^2}\) is zero, the points \(x = -5\) and \(x = 5\) are also critical.
Absolute Extrema
Absolute extrema refer to the absolute maximum or minimum values that a function attains on a specific interval. Unlike local extrema, which might occur in smaller open intervals within a function's domain, absolute extrema evaluate the function across a broader interval and include endpoints.
  • To determine absolute extrema:
  • Evaluate the function at each critical point and at the endpoints of the interval.
  • Compare these values to find the greatest and smallest ones.
For \(f(x) = \sqrt{25 - x^2}\) over \([-5, 5]\), we evaluated \(f(-5) = 0\), \(f(0) = 5\), and \(f(5) = 0\). The absolute maximum value is 5, and the absolute minimum is 0.
Differentiation
Differentiation is the process of finding a derivative, which represents the rate of change of a function with respect to its variable. Derivatives help identify critical points and can also be useful in optimizing real-world scenarios where you need to find maximum or minimum values.
  • A derivative can be found using basic rules like the power rule, product rule, quotient rule, or the chain rule.
  • For composite functions, the chain rule is particularly useful.
In this exercise, differentiating \(f(x) = (25 - x^2)^{1/2}\) gave us \(f'(x) = \frac{-x}{\sqrt{25 - x^2}}\), using the chain rule due to the composition of a square root function.
Chain Rule
The chain rule is a technique in calculus used to differentiate composite functions. It's like a two-step process that allows you to deal with functions nested within one another.
  • The general formula for the chain rule is: if you have a composite function, \(y = g(f(x))\), then its derivative is \(y' = g'(f(x)) \cdot f'(x)\).
  • This approach simplifies differentiation tasks that might otherwise be very complex.
In this specific problem, the function \(f(x) = (25 - x^2)^{1/2}\) is a composite function. By applying the chain rule, we first differentiated the outer function, \((...)^{1/2}\), then multiplied it by the derivative of the inner function, \((25 - x^2)\). This led us to the derivative \(f'(x) = \frac{-x}{\sqrt{25 - x^2}}\).

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Most popular questions from this chapter

Sketch the graphs of the following functions indicating any relative and absolute extrema, points of inflection, intervals on which the function is increasing, decreasing, concave upward, or concave downward. $$ f(x)=\frac{x+4}{x-4} $$

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Find \(\frac{d^{2} y}{d x^{2}}\) if \(y=\cos (2 x)+3 x^{2}-1\).

A function \(f\) is continuous on the interval [-2,5] with \(f(-2)=10\) and \(f\) \((5)=6\) and the following properties: $$ \begin{array}{c|c|c|c|c|c} \text { INTERVALS } & (-2,1) & X=1 & (1,3) & X=3 & (3,5) \\ \hline f^{\prime} & \+ & 0 & \- & \text { undefined } & \+ \\ \hline f^{\prime \prime} & \- & 0 & \- & \text { undefined } & + \end{array} $$ (a) Find the intervals on which \(f\) is increasing or decreasing. (b) Find where \(f\) has its absolute extrema. (c) Find where \(f\) has points of inflection. (d) Find the intervals where \(f\) is concave upward or downward. (e) Sketch a possible graph of \(f\).

Find \(\frac{d y}{d x}\) if \(\left(x^{2}+y^{2}\right)^{2}=10 x y\).
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