Over what open interval does the formula F(x)= int from 0 to x of dt/t represent an anti derivative of f(x) =1/x?

Find a point where the graph of F crosses the x axis.

To determine the open interval over which the formula F(x) represents an antiderivative of f(x), we need to check for two conditions:

1. Continuity: The function f(x) = 1/x should be continuous over the interval.
2. Differentiability: The function f(x) = 1/x should be differentiable over the interval, except possibly at isolated points where f(x) is not defined (such as x = 0).

First, let's consider continuity. Since the function f(x) = 1/x is not defined for x = 0, we need to find the interval where this function is continuous. Any open interval that does not include x = 0 will satisfy this condition. Thus, the open interval for the function f(x) = 1/x is (-∞, 0) ∪ (0, ∞).

Next, let's consider differentiability. Since the function f(x) = 1/x is defined and differentiable for all x ≠ 0, except at x = 0, the open interval for the function f(x) = 1/x as a differentiable function is (-∞, 0) ∪ (0, ∞).

Now, let's find a point where the graph of F crosses the x-axis. For F(x) = ∫(0 to x) 1/t dt, we can find the x-value where F(x) crosses the x-axis by setting F(x) equal to zero and solving for x:

0 = ∫(0 to x) 1/t dt

To evaluate the integral, we can use the logarithmic rule, which states that the integral of 1/t with respect to t is ln|t| + C. Applying this rule:

0 = ln|x| + C

Since C is an arbitrary constant, we can leave it as is or choose a specific value if desired.

To find a point where the graph of F crosses the x-axis, we set F(x) = 0:

ln|x| + C = 0

Now, we can exponentiate both sides to eliminate the natural logarithm:

e^(ln|x|+C) = e^0

|x|e^C = 1

Since e^C is a positive constant, we can rewrite the equation as:

|x| = 1

There are two solutions for |x| = 1: x = 1 and x = -1. Therefore, the graph of F crosses the x-axis at x = 1 and x = -1.

In summary:
- The open interval over which F(x) represents an antiderivative of f(x) = 1/x is (-∞, 0) ∪ (0, ∞).
- The graph of F crosses the x-axis at x = 1 and x = -1.