Question

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

f(x) = x^2 − 4x
------------ divided by
x^2 − 16


x = 4.1, 4.05, 4.01, 4.001, 4.0001,
3.9, 3.95, 3.99, 3.999, 3.9999

Answer 1

The function
`f(x) = (x^(2)-4\cdot x)/(x^(2)-16)` has the following set of solutions:

`f(4.1) = 0.506173`,
`f(4.05) = 0.503106`,
`f(4.01) = 0.500624`,
`f(4.001) = 0.500062`,
`f(4.0001) = 0.500006`,
`f(4.0001) = 0.500006`

`f(3.9) = 0.493671`,
`f(3.95) = 0.496855`,
`f(3.99) = 0.499374`,
`f(3.999) = 0.499937`,
`f(3.9999) = 0.499994`

Explanation:

Let be
`f(x) = (x^(2)-4\cdot x)/(x^(2)-16)`, we proceed to simplify the expression by Algebraic means:

1)
`(x^(2)-4\cdot x)/(x^(2)-16)` Given

2)
`(x\cdot (x-4))/((x-4)\cdot (x+4))` Associative, commutative and distributive properties/
`a^(2)-b^(2) = (a+b)\cdot (a - b)`

3)
`(x)/(x + 4)` Commutative, associative and modulative properties/Existence of multiplicative inverse/Result

Now we evaluate the function for each value:

`x = 4.1`

`f(4.1) = (4.1)/(4.1+4)`

`f(4.1) = 0.506173`

`x = 4.05`

`f(4.05) = (4.05)/(4.05 + 4)`

`f(4.05) = 0.503106`

`x = 4.01`

`f(4.01) = (4.01)/(4.01 + 4)`

`f(4.01) = 0.500624`

`x = 4.001`

`f(4.001) = (4.001)/(4.001+4)`

`f(4.001) = 0.500062`

`x = 4.0001`

`f(4.0001) = (4.0001)/(4.0001 + 4)`

`f(4.0001) = 0.500006`

`x = 3.9`

`f(3.9) = (3.9)/(3.9+4)`

`f(3.9) = 0.493671`

`x = 3.95`

`f(3.95) = (3.95)/(3.95+4)`

`f(3.95) = 0.496855`

`x = 3.99`

`f(3.99) = (3.99)/(3.99+4)`

`f(3.99) = 0.499374`

`x = 3.999`

`f(3.999) = (3.999)/(3.999 + 4)`

`f(3.999) = 0.499937`

`x = 3.9999`

`f(3.9999) = (3.9999)/(3.9999 + 4)`

`f(3.9999) = 0.499994`