Find f(g(x)) f(x)=2x^2+x-3 and g(x)=x-1

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

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

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

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

Step 3

Remove parentheses.

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

Step 4

Simplify each term.

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

$f (x - 1) = 2 ((x - 1) (x - 1)) + x - 1 - 3$

Expand $(x - 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$f (x - 1) = 2 (x (x - 1) - 1 (x - 1)) + x - 1 - 3$

Apply the distributive property.

$f (x - 1) = 2 (x \cdot x + x \cdot - 1 - 1 (x - 1)) + x - 1 - 3$

Apply the distributive property.

$f (x - 1) = 2 (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1) + x - 1 - 3$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $- 1$ to the left of $x$ .

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

Rewrite $- 1 x$ as $- x$ .

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

Rewrite $- 1 x$ as $- x$ .

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

Multiply $- 1$ by $- 1$ .

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

Subtract $x$ from $- x$ .

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

Apply the distributive property.

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

Simplify.

Multiply $- 2$ by $2$ .

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

Multiply $2$ by $1$ .

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

Step 5

Simplify by adding terms.

Add $- 4 x$ and $x$ .

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

Simplify by subtracting numbers.

Subtract $1$ from $2$ .

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

Subtract $3$ from $1$ .

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

Find f(g(x)) f(x)=2x^2+x-3 and g(x)=x-1

Related Questions

Support LCMGCF.com by disabling your adblocker.