Question

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

Answer 1

Answer:

The function $f(x) = \frac{x^{2}-4\cdot x}{x^{2}-16}$ has the following set of solutions:

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

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

Step-by-step explanation:

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

1) $\frac{x^{2}-4\cdot x}{x^{2}-16}$ Given

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

3) $\frac{x}{x + 4}$ Commutative, associative and modulative properties/Existence of multiplicative inverse/Result

Now we evaluate the function for each value:

$x = 4.1$

$f(4.1) = \frac{4.1}{4.1+4}$

$f(4.1) = 0.506173$

$x = 4.05$

$f(4.05) = \frac{4.05}{4.05 + 4}$

$f(4.05) = 0.503106$

$x = 4.01$

$f(4.01) = \frac{4.01}{4.01 + 4}$

$f(4.01) = 0.500624$

$x = 4.001$

$f(4.001) = \frac{4.001}{4.001+4}$

$f(4.001) = 0.500062$

$x = 4.0001$

$f(4.0001) = \frac{4.0001}{4.0001 + 4}$

$f(4.0001) = 0.500006$

$x = 3.9$

$f(3.9) = \frac{3.9}{3.9+4}$

$f(3.9) = 0.493671$

$x = 3.95$

$f(3.95) = \frac{3.95}{3.95+4}$

$f(3.95) = 0.496855$

$x = 3.99$

$f(3.99) = \frac{3.99}{3.99+4}$

$f(3.99) = 0.499374$

$x = 3.999$

$f(3.999) = \frac{3.999}{3.999 + 4}$

$f(3.999) = 0.499937$

$x = 3.9999$

$f(3.9999) = \frac{3.9999}{3.9999 + 4}$

$f(3.9999) = 0.499994$