Find g(f(x)) f(x)=4x+1 g(x)=x^2-5

$f (x) = 4 x + 1$ $g (x) = x^{2} - 5$

Step 1

Set up the composite result function.

$g (f (x))$

Step 2

Evaluate $g (4 x + 1)$ by substituting in the value of $f$ into $g$ .

$g (4 x + 1) = {(4 x + 1)}^{2} - 5$

Step 3

Simplify each term.

Rewrite ${(4 x + 1)}^{2}$ as $(4 x + 1) (4 x + 1)$ .

$g (4 x + 1) = (4 x + 1) (4 x + 1) - 5$

Expand $(4 x + 1) (4 x + 1)$ using the FOIL Method.

Apply the distributive property.

$g (4 x + 1) = 4 x (4 x + 1) + 1 (4 x + 1) - 5$

Apply the distributive property.

$g (4 x + 1) = 4 x (4 x) + 4 x \cdot 1 + 1 (4 x + 1) - 5$

Apply the distributive property.

$g (4 x + 1) = 4 x (4 x) + 4 x \cdot 1 + 1 (4 x) + 1 \cdot 1 - 5$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$g (4 x + 1) = 4 \cdot (4 x \cdot x) + 4 x \cdot 1 + 1 (4 x) + 1 \cdot 1 - 5$

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

Move $x$ .

$g (4 x + 1) = 4 \cdot (4 (x \cdot x)) + 4 x \cdot 1 + 1 (4 x) + 1 \cdot 1 - 5$

Multiply $x$ by $x$ .

$g (4 x + 1) = 4 \cdot (4 x^{2}) + 4 x \cdot 1 + 1 (4 x) + 1 \cdot 1 - 5$

Multiply $4$ by $4$ .

$g (4 x + 1) = 16 x^{2} + 4 x \cdot 1 + 1 (4 x) + 1 \cdot 1 - 5$

Multiply $4$ by $1$ .

$g (4 x + 1) = 16 x^{2} + 4 x + 1 (4 x) + 1 \cdot 1 - 5$

Multiply $4 x$ by $1$ .

$g (4 x + 1) = 16 x^{2} + 4 x + 4 x + 1 \cdot 1 - 5$

Multiply $1$ by $1$ .

$g (4 x + 1) = 16 x^{2} + 4 x + 4 x + 1 - 5$

Add $4 x$ and $4 x$ .

$g (4 x + 1) = 16 x^{2} + 8 x + 1 - 5$

Step 4

Subtract $5$ from $1$ .

$g (4 x + 1) = 16 x^{2} + 8 x - 4$