Find the tangent line approximation to 1/x when x=1
How do i find tangent line approximations??? my prof hasn't gone over this and i don't understand how to do it.
y = 1/x
dy/dx = -1/x^2 = slope
at x = 1, slope = -1/1^2 = -1
find the line with that slope through that point (1,1)
1 = -1 *1 + b
b = 2
so line
y = -x + 2
The function is 1/x. There are a number of approximations, but I suspect you are using this one...
f(x+h)-f(x) / h
(1/(x+h)-1/x )/ h
so let h be some small increment, say .05
(1/1.05 - 1/1)/.05= you do it, that is the slope of the tangent line approximation.
thanks!!
To find the tangent line approximation to a function at a specific point, you will need to use calculus and the concept of derivatives. Here's a step-by-step explanation of how to find the tangent line approximation to the function 1/x when x=1:
1. Start by finding the derivative of the function. In this case, the derivative of 1/x is obtained using the power rule for differentiation: f'(x) = -1/x^2.
2. Substitute the given x-value (x=1) into the derivative to find the slope of the tangent line at that point. In this case, f'(1) = -1/(1^2) = -1.
3. Now that you know the slope of the tangent line, you need the y-coordinate of the point of tangency. Plug the given x-value (x=1) back into the original function (1/x) to find f(1). In this case, f(1) = 1/1 = 1.
4. Use the point-slope form of a line (y - y1 = m(x - x1)) to write the equation of the tangent line using the slope (m) and the point of tangency (x1, y1). Plugging in the values we found earlier, you get y - 1 = -1(x - 1).
5. Simplify the equation to obtain the final tangent line approximation. Distribute and rearrange terms to get y = -x + 2.
So, the equation of the tangent line approximation to 1/x when x=1 is y = -x + 2.