The slope of the tangent line at the point x = a of the function f(x) is f'(a).
We are given the function:
[tex]f(x)=\frac{1}{x}[/tex]
Computing the first derivative:
[tex]f^{\prime}(x)=-\frac{1}{x^2}[/tex]
The slope of the tangent line at (1, 1), that is, where x = 1 is:
[tex]f^{\prime}(1)=-\frac{1}{1^2}=-1[/tex]
The tangent line and the normal line are perpendicular to each other. If their respective slopes are m1 and m2, then:
[tex]m_1\cdot m_2=-1[/tex]
We have calculated m1 = -1, calculate m2:
[tex]m_2=-\frac{1}{m_1}=-\frac{1}{-1}=1[/tex]
Now we know the slope of the normal line. We need to find its equation. Use the point-slope formula:
y - k = m(x - h)
Where m is the known slope and (h, k) is a point of the line. We are given the point (1, 1), thus:
y - 1 = 1(x - 1) = x - 1
Adding 1:
y = x
Answer: D. x
