Question

Evaluate the function f(x) at the given numbers (correct to six decimal places). F(x) = x2 − 5x x2 − x − 20 , x = 5.5, 5.1, 5.05, 5.01, 5.005, 5.001, 4.9, 4.95, 4.99, 4.995, 4.999

Answer 1

we are given

$f(x)=\frac{x^2-5x}{x^2-x-20}$

Firstly, we will simplify it

$f(x)=\frac{x(x-5)}{(x-5)(x+4)}$

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

At x=5.5:

we can plug x=5.5

$f(5.5)=\frac{5.5}{(5.5+4)}$

$f(5.5)=0.578947$

At x=5.1:

we can plug x=5.1

$f(5.1)=\frac{5.1}{(5.1+4)}$

$f(5.1)=0.56043956$

At x=5.05:

we can plug x=5.05

$f(5.05)=\frac{5.05}{(5.05+4)}$

$f(5.05)=0.558011$

At x=5.01:

we can plug x=5.01

$f(5.01)=\frac{5.01}{(5.01+4)}$

$f(5.01)=0.5560488$

At x=5.005:

we can plug x=5.005

$f(5.005)=\frac{5.005}{(5.005+4)}$

$f(5.005)=0.5558023$

At x=5.001:

we can plug x=5.001

$f(5.001)=\frac{5.001}{(5.001+4)}$

$f(5.001)=0.5556049$

At x=4.9:

we can plug x=4.9

$f(4.9)=\frac{4.9}{(4.9+4)}$

$f(4.9)=0.5505617$

At x=4.95:

we can plug x=4.95

$f(4.95)=\frac{4.95}{(4.95+4)}$

$f(4.95)=0.5530726$

At x=4.99:

we can plug x=4.99

$f(4.99)=\frac{4.99}{(4.99+4)}$

$f(4.99)=0.55506117$

At x=4.995:

we can plug x=4.995

$f(4.995)=\frac{4.995}{(4.995+4)}$

$f(4.995)=0.55503085$

At x=4.999:

we can plug x=4.999

$f(4.999)=\frac{4.999}{(4.999+4)}$

$f(4.999)=0.55550616$