Question

Use linear approximation, i.e. the tangent line, to approximate 1/0.502 as follows: Let f(x)=1/x and find the equation of the tangent line to f(x) at a "nice" point near 0.502. Then use this to approximate 1/0.502.

Answer 1

so a tangent line is parallel at the point (0.502), and inorder to find the tangent line we need the slope, m. m = (y2-y1) / (x2-x1)
so lets pick 2 points near our point of 0.502, or inother words, lets do linear approximation.
so solving 1/x where x1 = 0.501 and then do x2 = 0.503
now we get two points
(0.501, 1.9960)
(0.503, 1.9881)

can you find the slope from here with these two points then find the equation of the tangent line?

Answer 2

I got -3.95 as the slop

Answer 3

yes, now use the y=mx+b formula. you now have m, but now you need x and y, so use the x point given you, 0.502 and solve for y so, 1/0.502 = y.
now you have m, x, and y, now you can solve for b in the y=mx+b equation

Answer 4

What do I do once I get b?

Answer 5

rewrite the equation of y=mx+b, and replace m and b with the values you found

Answer 6

so,
f(x) = -3.95x -1.0046

Answer 7

Okay, I got that

Answer 8

is there anything else you need help with?

Answer 9

I'm not sure where to go after that...