Calculus problem with Linear Approximation

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 in slope-intercept form to f(x) at a "nice" point near  0.502 . Then use this to approximate 1/0.502.

_________

Answer 1

Choose a nice point where x = 1/2 (0.5). When x = 1/2, y = 1/(1/2) = 2. So a point on the curve is (1/2, 2).

To determine the slope of the tangent line at that point, find f'(1/2).

f(x) = x^-1

f'(x) = -x^-2 = -1/x²

f'((1/2) = -1/(1/2)² = -1/(1/4) = -4

So we have a point: (1/2, 2) and a slope (-4) for the tangent line at that point.

y - 2 = -4(x - 1/2)

y - 2 = -4x + 2

y = -4x + 4

L(x) = -4x + 4

L(.502) = -4(0.502) + 4 = 1.992

This is an approximation for the exact value:

f(0.502) = 1/0.502 ≈ 1.99203187251

Here is a Desmos graph to help picture the above:

desmos.com/calculator/ycsd9kfstz

Here is the same graph honed in on f(0.502) and L(0.502). The difference between the curve 1/x and its tangent line is barely perceptible.

desmos.com/calculator/ldex2xohir