Question

For f(x) =1/x^2-3 find: (a) f(3) (3 points) (b) f(2+h) (3 points)

Answer 1

Answer:

$f(3) = \frac{-26}{9}$

$f(2 + h) = \frac{- 11 - 12h - 12^2}{4 + 4h + h^2}$

Step-by-step explanation:

Given

$f(x) = \frac{1}{x^2} - 3$

Required

Find f(3) and f(2 + h)

Solving f(3)

Substitute 3 for x in $f(x) = \frac{1}{x^2} - 3$

$f(3) = \frac{1}{3^2} - 3$

$f(3) = \frac{1}{9} - 3$

Take LCM

$f(3) = \frac{1 - 27}{9}$

$f(3) = \frac{-26}{9}$

Solving f(2 + h)

Substitute 2 + h for x in $f(x) = \frac{1}{x^2} - 3$

$f(2 + h) = \frac{1}{(2 + h)^2} - 3$

$f(2 + h) = \frac{1}{(2 + h)(2 + h)} - 3$

$f(2 + h) = \frac{1}{4 + 2h + 2h + h^2} - 3$

$f(2 + h) = \frac{1}{4 + 4h + h^2} - 3$

Take LCM

$f(2 + h) = \frac{1- 3(4 + 4h + h^2)}{4 + 4h + h^2}$

$f(2 + h) = \frac{1- 12 - 12h - 12^2}{4 + 4h + h^2}$

$f(2 + h) = \frac{- 11 - 12h - 12^2}{4 + 4h + h^2}$