1. The problem is to determine which of the functions \(f(x) = \frac{x^2 - 1}{x}\), \(f(x) = x \sqrt{x^2 + 1}\), or \(f(x) = x^2 - 1\) intersects the x-axis at \(x=1\) and is perpendicular to the base point at that intersection. 2. First, find which functions intersect the x-axis at \(x = 1\). The x-axis intersects where \(f(x) = 0\). 3. For \(f(x) = \frac{x^2 - 1}{x}\): \[f(1) = \frac{1^2 - 1}{1} = \frac{0}{1} = 0\] So it intersects at \(x=1\). 4. For \(f(x) = x \sqrt{x^2 + 1}\): \[f(1) = 1 \times \sqrt{1 + 1} = \sqrt{2} \ne 0\] No intersection at \(x=1\). 5. For \(f(x) = x^2 - 1\): \[f(1) = 1 - 1 = 0\] So it intersects at \(x=1\). 6. Now check the slopes (derivatives) of the functions at \(x=1\) to find which is perpendicular to the x-axis (y=0), meaning the slope should be infinite (vertical tangent) or the function's tangent line is perpendicular to the base. 7. Derivative of \(f(x) = \frac{x^2 - 1}{x} = x - \frac{1}{x}\): \[f'(x) = 1 + \frac{1}{x^2}\] So, \[f'(1) = 1 + 1 = 2\] 8. Derivative of \(f(x) = x \sqrt{x^2 + 1}\): Use product rule and chain rule: \[f'(x) = \sqrt{x^2 + 1} + x \times \frac{1}{2} (x^2 + 1)^{-1/2} 2x = \sqrt{x^2 + 1} + \frac{x^2}{\sqrt{x^2 + 1}}\] At \(x=1\): \[f'(1) = \sqrt{2} + \frac{1}{\sqrt{2}} = \sqrt{2} + \frac{1}{\sqrt{2}} = \frac{2 + 1}{\sqrt{2}} = \frac{3}{\sqrt{2}}\] 9. Derivative of \(f(x) = x^2 - 1\): \[f'(x) = 2x\] So, \[f'(1) = 2\] 10. Since the x-axis is horizontal with slope \(0\), a function is perpendicular to it if its slope is undefined (vertical) at that point. None of these derivatives is infinite at \(x=1\). 11. However, the problem might mean that the function’s graph at \(x=1\) is perpendicular to the base point’s tangent on the x-axis, so the slope of the function should be the negative reciprocal of the x-axis slope \(= 0\). The negative reciprocal of 0 is undefined, meaning vertical tangent. 12. None of the functions has an infinite slope at \(x=1\), so none is perpendicular to the x-axis at \(x=1\). 13. But considering practical cases, the function \(f(x) = \frac{x^2 - 1}{x} = x - \frac{1}{x}\) has a vertical asymptote at \(x=0\) but at \(x=1\) it has slope 2, not perpendicular. 14. The function \(f(x) = x^2 - 1\) has slope 2 at \(x=1\) and passes through zero there. 15. Function \(f(x) = x \sqrt{x^2 + 1}\) doesn't intersect the x-axis at \(x=1\). Final conclusion: Among given functions, only \(f(x) = \frac{x^2-1}{x}\) and \(f(x) = x^2 -1\) intersect the x-axis at \(x=1\), but neither has a perpendicular slope (vertical) at \(x=1\). Both have slope 2 there. Hence, from intersection perspective at \(x=1\), options A and C qualify for intersection, but none is perpendicular to the x-axis there.