Question

Use linear approximation, i.e. the tangent line, to approximate 1/0.201 as follows: Let f (x) = 1/x and find the equation of the tangent line to f(x) at a "nice" point near 0.201 . Then use this to approximate 1/0.201.

Answer 1

Answer:

f(0.201) ≈ 4.975  

Step-by-step explanation:

1. Calculate the position of the point and the slope of the tangent

$f(x) = \dfrac{1}{x}\\\\f'(x) = -\dfrac{1}{x^{2}}$

When x = 0.2

$f(0.2) = \dfrac{1}{0.2} = 5\\\\f'(0.2) = - \dfrac{1}{0.2^{2}} = -\dfrac{1}{0.04} = -25$

So, the point is at (0.2, 5) and the slope of the tangent is -25.

2. Calculate the equation of the tangent line

y = mx+ b

Substitute the coordinates of the point

5 = -25×0.2 + b

5 = -5 + b

b = 10

The equation of the tangent line is

y = -25x + 10

3. Calculate y when x = 0.201

y = -25×0.201 + 10 = -5.025 + 10 = 4.975

4. How good is the approximation?

The actual value of f(0.201) is 4.975 124 …

$\text{Percent error} = \dfrac{0.000124}{4.975124} \times 100 \, \% = \mathbf{0.0025 \, \%}$