Finite Math Examples

f (x) = 1 - x

$f (x) = 1 - x$ ,

g (x) = 2 x^{2} + x + 5

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

Step 1

Set up the composite result function.

f (g (x))

$f (g (x))$

Step 2

Evaluate

f (2 x^{2} + x + 5)

$f (2 x^{2} + x + 5)$ by substituting in the value of

g

$g$ into

f

$f$ .

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

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

Step 3

Simplify each term.

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Step 3.1

Apply the distributive property.

f (2 x^{2} + x + 5) = 1 - (2 x^{2}) - x - 1 \cdot 5

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

Step 3.2

Simplify.

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Step 3.2.1

Multiply

2

$2$ by

- 1

$- 1$ .

f (2 x^{2} + x + 5) = 1 - 2 x^{2} - x - 1 \cdot 5

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

Step 3.2.2

Multiply

- 1

$- 1$ by

5

$5$ .

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

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

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

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

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

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

Step 4

Subtract

5

$5$ from

1

$1$ .

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

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

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

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

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