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

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

Step 1

Set up the composite result function.

$g (f (x))$

Step 2

Evaluate $g (2 x^{2})$ by substituting in the value of $f$ into $g$ .

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

Step 3

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

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

Step 4

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Step 5

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

Move $x^{2}$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $2$ and $2$ .

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

Multiply $2$ by $2$ .

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

Multiply $2$ by $1$ .

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

Multiply $2 x^{2}$ by $1$ .

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

Multiply $1$ by $1$ .

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

Add $2 x^{2}$ and $2 x^{2}$ .

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