Question

(a) Over what open interval does the formula \(F(x)=\int_{1}^{x} \frac{1}{t^{2}-9} d t\) represents an antiderivative of \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}^{2}-9}\) ? (b) Find a point where the graph of \(\mathrm{F}\) crosses the \(x\)-axis.

Short answer

Answer: The given formula represents an antiderivative of f(x) on the open interval (1,3)∪(3,∞). There is no point where the graph of F crosses the x-axis within these intervals.

Step-by-step solution

Determine the Integrand's Domain

The given integrand is \(f(x)=\frac{1}{x^2-9}\). We need to find the domain for this function, which is the set of x values for which the function exists. The function can be undefined only when the denominator is zero. \(x^2-9=0\) when \(x^2 = 9\), so \(x = \pm3\). Therefore, the domain of the integrand is \((-\infty, -3)\cup(-3, 3)\cup(3, \infty)\).

Determine the Interval of Antiderivative

Now that we know the domain of the integrand, we can determine the interval over which the given formula represents an antiderivative for f(x). When we try to integrate the function, we need to take into account the points where the function is discontinuous. In this case, there are two such points: \(x = -3\) and \(x = 3\). Given that $F(x)=\int_{1}^{x} \frac{1}{t^{2}-9} d t\(, we can see that the integral is valid for x values greater than 1. Therefore, the open interval for which F(x) is an antiderivative of f(x) is \)(1,3)\cup(3,\infty)$.

Find the \(x\)-axis Intersection Point

To find a point where the graph of F crosses the x-axis, we need to determine when \(F(x) = 0\). Since $F(x)=\int_{1}^{x} \frac{1}{t^{2}-9} d t\(, we are looking for the value of x where the integral \)\int_{1}^{x} \frac{1}{t^{2}-9} d t = 0\(. For this to happen, the area under the curve of the function within the interval (1, x) should be zero (i.e. the positive and negative areas should cancel each other out). Looking at the function, it is clear that the function is not symmetric, and it is not clear if there is an x value within the interval \)(1,3)\cup(3,\infty)$ for which this condition will be satisfied. Thus, there is no point where the graph of F crosses the x-axis within the given intervals.