Determine the equation of the tangent to f(x)= (1/x)+x at the point (1,2)
f' = -1/x^2 + 1 = (x^2-1)/x^2
f'(1) = 0
so, using the point-slope form,
y-2 = 0(x-1)
or,
y=2
http://www.wolframalpha.com/input/?i=plot+y%3D1%2Fx+%2B+x,+y%3D2,+0+%3C%3D+x+%3C%3D+2
To determine the equation of the tangent line to the curve at a given point, you need to find the derivative of the function and evaluate it at that point. This will give you the slope of the tangent line. Once you have the slope, you can use the point-slope form of a linear equation to find the equation of the tangent line.
Let's go through the steps:
Step 1: Find the derivative of the function f(x).
The function f(x) = 1/x + x can be rewritten as f(x) = x^(-1) + x. To find the derivative, you can use the power rule and the sum rule of differentiation.
The derivative of x^(-1) is -1x^(-2) which simplifies to -1/x^2.
The derivative of x is 1.
So, the derivative of f(x) is: f'(x) = -1/x^2 + 1.
Step 2: Evaluate the derivative at the given point (1,2).
Plug x = 1 into the derivative f'(x):
f'(1) = -1/1^2 + 1 = -1 + 1 = 0.
The slope of the tangent line at the point (1, 2) is 0.
Step 3: Use the point-slope form to find the equation of the tangent line.
The point-slope form of a linear equation is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.
Plugging in the coordinates of the given point (1,2) and the slope (m=0), we get:
y - 2 = 0(x - 1).
Since the slope is 0, the equation simplifies to:
y - 2 = 0.
Therefore, the equation of the tangent line to f(x) = 1/x + x at the point (1, 2) is y = 2.