Question

Evaluate the function f(x) at the given numbers (correct to six decimal places).

Answer 1

Answer:

$f(2.1) = 0.512195$

$f(2.05) = 0.506173$

$f(2.01)=0.501247$

$f(2.001) = 0.500125$

$f(2.0001) = 0.500012$

$f(1.9) = 0.487179$

$f(1.95) = 0.493671$

$f(1.99) = 0.498747$

$f(1.999) = 0.499875$

$f(1.9999) = 0.499987$

Step-by-step explanation:

Given

$f(x) = \frac{x^2 - 2x}{x^2 - 4}$

Solve for f(x) for all given values of x

First, we need to simplify f(x)

$f(x) = \frac{x^2 - 2x}{x^2 - 4}$

$f(x) = \frac{x(x - 2)}{x^2 - 2^2}$

$f(x) = \frac{x(x - 2)}{(x- 2)(x + 2)}$

$f(x) = \frac{x}{x + 2}$

When $x = 2.1$

$f(x) = \frac{x}{x + 2}$

$f(2.1) = \frac{2.1}{2.1 + 2}$

$f(2.1) = \frac{2.1}{4.1}$

$f(2.1) = 0.512195$

When $x = 2.05$

$f(x) = \frac{x}{x + 2}$

$f(2.05) = \frac{2.05}{2 + 2.05}$

$f(2.05) = \frac{2.05}{4.05}$

$f(2.05) = 0.506173$

When $x = 2.01$

$f(x) = \frac{x}{x + 2}$

$f(2.01)=\frac{2.01}{2.01 +2}$

$f(2.01)=\frac{2.01}{4.01}$

$f(2.01)=0.501247$

When $x = 2.001$

$f(x) = \frac{x}{x + 2}$

$f(2.001) = \frac{2.001}{2.001 +2}$

$f(2.001) = \frac{2.001}{4.001}$

$f(2.001) = 0.500125$

When $x = 2.0001$

$f(x) = \frac{x}{x + 2}$

$f(2.0001) = \frac{2.0001}{2.0001 + 2}$

$f(2.0001) = \frac{2.0001}{4.0001}$

$f(2.0001) = 0.500012$

When $x = 1.9$

$f(x) = \frac{x}{x + 2}$

$f(1.9) = \frac{1.9}{1.9 + 2}$

$f(1.9) = \frac{1.9}{3.9}$

$f(1.9) = 0.487179$

When $x = 1.95$

$f(x) = \frac{x}{x + 2}$

$f(1.95) = \frac{1.95}{1.95 + 2}$

$f(1.95) = \frac{1.95}{3.95}$

$f(1.95) = 0.493671$

When $x = 1.99$

$f(x) = \frac{x}{x + 2}$

$f(1.99) = \frac{1.99}{1.99 + 2}$

$f(1.99) = \frac{1.99}{3.99}$

$f(1.99) = 0.498747$

$f(x) = \frac{x}{x + 2}$

When $x = 1.999$

$f(x) = \frac{x}{x + 2}$

$f(1.999) = \frac{1.999}{1.999 + 2}$

$f(1.999) = \frac{1.999}{3.999}$

$f(1.999) = 0.499875$

When x = 1.9999

$f(x) = \frac{x}{x + 2}$

$f(1.9999) = \frac{1.9999}{1.9999 + 2}$

$f(1.9999) = \frac{1.9999}{3.9999}$

$f(1.9999) = 0.499987$

Note that all values of f(x) are approximated to 6 decimal places