Find f(g(x)) f(x)=x^2-4 , g(x)=2x+1

$f (x) = x^{2} - 4$ , $g (x) = 2 x + 1$

Set up the composite result function.

$f (g (x))$

Evaluate $f (2 x + 1)$ by substituting in the value of $g$ into $f$ .

$f (2 x + 1) = {(2 x + 1)}^{2} - 4$

Simplify each term.

Rewrite ${(2 x + 1)}^{2}$ as $(2 x + 1) (2 x + 1)$ .

$f (2 x + 1) = (2 x + 1) (2 x + 1) - 4$

Expand $(2 x + 1) (2 x + 1)$ using the FOIL Method.

Apply the distributive property.

$f (2 x + 1) = 2 x (2 x + 1) + 1 (2 x + 1) - 4$

Apply the distributive property.

$f (2 x + 1) = 2 x (2 x) + 2 x \cdot 1 + 1 (2 x + 1) - 4$

Apply the distributive property.

$f (2 x + 1) = 2 x (2 x) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1 - 4$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$f (2 x + 1) = 2 \cdot (2 x \cdot x) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1 - 4$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$f (2 x + 1) = 2 \cdot (2 (x \cdot x)) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1 - 4$

Multiply $x$ by $x$ .

$f (2 x + 1) = 2 \cdot (2 x^{2}) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1 - 4$

Multiply $2$ by $2$ .

$f (2 x + 1) = 4 x^{2} + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1 - 4$

Multiply $2$ by $1$ .

$f (2 x + 1) = 4 x^{2} + 2 x + 1 (2 x) + 1 \cdot 1 - 4$

Multiply $2 x$ by $1$ .

$f (2 x + 1) = 4 x^{2} + 2 x + 2 x + 1 \cdot 1 - 4$

Multiply $1$ by $1$ .

$f (2 x + 1) = 4 x^{2} + 2 x + 2 x + 1 - 4$

Add $2 x$ and $2 x$ .

$f (2 x + 1) = 4 x^{2} + 4 x + 1 - 4$

Subtract $4$ from $1$ .

$f (2 x + 1) = 4 x^{2} + 4 x - 3$