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Chapter 4: Problem 7

Verify that the hypotheses of the Mean-Value Theorem are satisfied on the given interval, and find all values of \(c\) in that interval that satisfy the conclusion of the theorem. $$ f(x)=\sqrt{25-x^{2}} ;[-5,3] $$

Short Answer

Expert verified
The value of \( c \) is \( \frac{4}{5} \).

Step by step solution

01

Verify Continuity

Check whether the function \( f(x) = \sqrt{25-x^2} \) is continuous on the closed interval \([-5, 3]\). Since it is a composition of continuous functions (a square root and a polynomial), and its domain restricts \( x^2 \leq 25 \), \( f(x) \) is continuous on \([-5, 3]\) because the endpoints satisfy \( -5 \leq x \leq 5 \).
02

Verify Differentiability

Check the differentiability of \( f(x) \) on the open interval \((-5, 3)\). The function \( f(x) = \sqrt{25-x^2} \) is differentiable on \(-5 < x < 5\), thus differentiable on the open interval \((-5, 3)\). It is smooth within the bounds of its domain except at the endpoints.
03

Apply the Mean Value Theorem

Since \( f(x) \) is continuous on \([-5, 3]\) and differentiable on \((-5, 3)\), the Mean Value Theorem is applicable. This states there exists at least one \( c \) in \((-5, 3)\) such that \( f'(c) = \frac{f(3) - f(-5)}{3 - (-5)} \).
04

Compute \( f(a) \) and \( f(b) \)

Calculate \( f(3) \) and \( f(-5) \). We have \( f(3) = \sqrt{25 - 3^2} = \sqrt{16} = 4 \) and \( f(-5) = \sqrt{25 - (-5)^2} = \sqrt{0} = 0 \). So, \( f(3) - f(-5) = 4 - 0 = 4 \).
05

Calculate \( \frac{f(b) - f(a)}{b - a} \)

Evaluate \( \frac{f(3) - f(-5)}{3 + 5} = \frac{4}{8} = \frac{1}{2} \). Thus, \( f'(c) = \frac{1}{2} \).
06

Find \( f'(x) \)

Compute the derivative \( f'(x) = \frac{d}{dx}(\sqrt{25-x^2}) \). Using the chain rule, \( f'(x) = \frac{-x}{\sqrt{25-x^2}} \).
07

Solve for \( c \)

Set \( f'(c) = \frac{1}{2} \), we have \( \frac{-c}{\sqrt{25-c^2}} = \frac{1}{2} \). Squaring both sides and simplifying, \( -2c = \sqrt{25-c^2} \) gives \( 4c^2 + 25 = c^2 \) and after solving it leads to \( c = \frac{4}{5} \).
08

Verify \( c \) in the Interval

Check that the solution \( c = \frac{4}{5} \) is within the interval \((-5, 3)\). Since \( \frac{4}{5} \approx 0.8 \), it clearly lies within the interval, confirming the result.

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Key Concepts

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

Continuity
Continuity is a fundamental requirement to apply the Mean Value Theorem. A function is continuous on an interval when there are no breaks, jumps, or holes over that stretch of the curve. For a function to be continuous on a closed interval \([-5, 3]\), it must be well-defined and seamless from one endpoint to the other.
  • The function \( f(x) = \sqrt{25-x^2} \) is based on a square root function of a quadratic, which are both continuous by nature.
  • However, we must also ensure that the expression \( 25-x^2 \) is non-negative throughout the interval so the square root is defined, meaning \( x^2 \leq 25 \).
This condition is met in \([-5, 3]\), as substituting the end values results in valid numbers for the square root function. Thus, \( f(x) \) is continuous on the interval.
Differentiability
Differentiability is another crucial criterion for applying the Mean Value Theorem. A function is differentiable at a point if it has a defined tangent line at that point. This implies that the function should not have any sharp corners or cusps.For \( f(x) = \sqrt{25-x^2} \) in the open interval \((-5, 3)\), check by:
  • Verifying that the function forms a smooth curve, except potentially at the endpoints. Differentiability is not necessarily guaranteed at endpoints of a closed interval.
  • Given the nature of \( \sqrt{25-x^2} \,\), it is smooth inside the open range \((-5, 5)\).
Thus, the function is differentiable in our considered range. Note that the differentiability has to be checked within the open interval and considers \( f(x) \'s\) overall smoothness which in this case, \(\sqrt{25-x^2}\), allows.
Square root function
The square root function \( f(x) = \sqrt{25-x^2} \) involves finding the non-negative square root of expressions within it. Key points about \( \sqrt{25-x^2} \,\):
  • This function represents a semicircle since the equation \( x^2 + y^2 = 25 \) describes a circle radius 5 centered at the origin. By setting \( y = \sqrt{25-x^2}\), it takes the upper half of that circle.
  • Understanding this geometric view aids in seeing why the function is bounded and continuous, as the circle is a continuous shape.
Therefore, within the interval \([-5, 3]\), \( \sqrt{25-x^2} \) successfully depicts a continuous and differentiable path due to its semicircle nature.
Derivative calculation
Calculating the derivative of a square root function, especially one like \( \sqrt{25-x^2} \), follows specific rules. Applying the chain rule:
  • The derivative \( f'(x) = \frac{d}{dx}(\sqrt{25-x^2}) \) involves understanding the derivative of a composite function:
  • Using the chain rule gives \( f'(x) = \frac{-x}{\sqrt{25-x^2}} \).
To apply the Mean Value Theorem, we find where this derivative equals the average rate of change over the interval. Thus, \( f'(c) = \frac{1}{2} \) and solving leads to \(-2c = \sqrt{25-c^2} \), which upon solving simplifies to \( c = \frac{4}{5} \). This value \( c \) is found well within the required range, confirming smooth differentiation allows for the theorem鈥檚 application.

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