Question

Discuss the continuity of the function on the closed interval. $$ \begin{array}{ll} \text { Function } & \text { Interval } \\ \hline g(x)=\sqrt{25-x^{2}} & {[-5,5]} \end{array} $$

Short answer

The function \(g(x)=\sqrt{25-x^{2}}\) is continuous on the closed interval \([-5,5]\).

Step-by-step solution

Check definition range of the function

First, we must check if all the real numbers in the given interval \([-5,5]\) are inside the definition range of the function. This means checking that the term under the square root, i.e., \(25 - x^2\), is always greater than or equal to zero in the given interval. For this, we solve the following inequality: \(25 - x^2 >= 0\). Solving this gives us the solution interval \([-5,5]\), which matches the interval given in the problem. This means there is no real number in the given interval that would make \(25 - x^2\) negative.

Verify continuity

Now that we have confirmed that the function is defined on the entire interval, the next step would be to check its continuity. The function \(g(x)=\sqrt{25-x^{2}}\) is a composition of two continuous functions: the square root function and the function \(25 - x^2\). Since a composition of continuous functions is, by definition, itself continuous, it follows that \(g(x) = \sqrt{25-x^{2}}\) is continuous wherever it is defined.

Final confirmation

We have seen that all real numbers in the given interval \([-5,5]\) are in the definition range of the function and that the function is continuous at every point in its domain. Thus, it is safe to conclude that the function is continuous on the closed interval \([-5,5]\).

Key concepts

Closed Interval

In mathematics, a closed interval is a range of numbers that includes both its endpoints. Formally, a closed interval is denoted by square brackets, as \[ [a, b] \], which signifies that all real numbers, including the endpoints \(a\) and \(b\), are part of the interval.

Understanding closed intervals is crucial when discussing the continuity of a function over a specific range. In the given exercise, the interval \[ [-5, 5] \] is a closed interval since it includes both endpoints, -5 and 5. When assessing whether a function such as \(g(x) = \sqrt{25-x^2}\) is continuous, we are specifically concerned with every point within this interval, emphasizing the inclusion of the endpoints.

Definition Range of a Function

The definition range, or simply the range, of a function consists of all the possible output values that the function can produce, based on the input values from its domain.

To determine if a function is continuous over a particular interval, we first verify if all the values in the interval fall within the domain of the function. This crucial step ensures that the function is capable of providing an output for all the input values in the interval. For example, in the exercise, the function \(g(x) = \sqrt{25 - x^2}\) is defined only if \(25 - x^2\) is greater than or equal to zero because the square root function requires non-negative inputs.

Square Root Function Continuity

The continuity of a square root function, like \(\sqrt{x}\), hinges on its input being non-negative. The square root function is continuous for all non-negative values of \(x\), as it smoothly maps to real number outputs without any interruptions or jumps.

In our specific case with \(g(x) = \sqrt{25 - x^2}\), we look into whether the expression within the square root, \(25 - x^2\), remains non-negative on the closed interval \[ [-5, 5] \]. It's determined that on this interval, the function does not produce any negative values inside the square root, establishing the domain of \(g(x)\) as \[ [-5, 5] \] and ensuring its continuity over this range.

Composition of Continuous Functions

The composition of continuous functions is a cornerstone concept in calculus. When two functions, say \(f(x)\) and \(h(x)\), are continuous individually and are combined to form a new function \(g(x) = f(h(x))\), the resulting function \(g(x)\) will also be continuous provided that the composition is well-defined over the domain in question.

For our exercise involving \(g(x) = \sqrt{25 - x^2}\), we see that it is a composition of the square root function and the quadratic function \(25 - x^2\), both of which are continuous in their respective domains. Therefore, their composition \(g(x)\) is continuous in its domain, which, as previously established, includes the closed interval \[ [-5, 5] \]. This illustrates that by understanding the continuous behavior of individual functions, we can ascertain the continuity of their composites across specific intervals.