Question

Evaluate the function f(x) at the given numbers (correct to six decimal places). f(x) = x2 − 3x x2 − 9 , x = 3.1, 3.05, 3.01, 3.001, 3.0001, 2.9, 2.95, 2.99, 2.999, 2.9999

Answer 1

Answer:

Values of f(x) for different values of x.        

Step-by-step explanation:

We are given the following function in the question:

$f(x) = \displaystyle\frac{x^2-3x}{x^2-9}$

We have to find the value of given function at the following x.

x = 3.1, 3.05, 3.01, 3.001, 3.0001, 2.9, 2.95, 2.99, 2.999, 2.9999

$f(x) = \displaystyle\frac{x^2-3x}{x^2-9}\\\\f(3.1) = \frac{(3.1)^2-3(3.1)}{(3.1)^2-9} = 0.508197\\\\f(3.05) = \frac{(3.05)^2-3(3.05)}{(3.05)^2-9} = 0.504132\\\\f(3.01) = \frac{(3.01)^2-3(3.01)}{(3.01)^2-9} = 0.500832\\\\f(3.001) = \frac{(3.001)^2-3(3.001)}{(3.001)^2-9} = 0.500083\\\\f(3.0001) = \frac{(3.0001)^2-3(3.0001)}{(3.0001)^2-9} = 0.500008\\\\f(2.9) = \frac{(2.9)^2-3(2.9)}{(2.9)^2-9} = 0.4915254\\\\f(2.95) = \frac{(2.95)^2-3(2.95)}{(2.95)^2-9} =0.495798$

$f(2.99) = \displaystyle\frac{(2.99)^2-3(2.99)}{(2.99)^2-9} =0.499165\\\\f(2.999) = \displaystyle\frac{(2.999)^2-3(2.999)}{(2.999)^2-9} = 0.499917\\\\f(2.9999) = \displaystyle\frac{(2.9999)^2-3(2.9999)}{(2.9999)^2-9} =0.499991$