Calculus Examples

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

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (2 x + 5)$ by substituting in the value of $g$ into $f$ .

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

Step 3

Rewrite ${(2 x + 5)}^{2}$ as $(2 x + 5) (2 x + 5)$ .

$f (2 x + 5) = (2 x + 5) (2 x + 5)$

Step 4

Expand $(2 x + 5) (2 x + 5)$ using the FOIL Method.

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Step 4.1

Apply the distributive property.

$f (2 x + 5) = 2 x (2 x + 5) + 5 (2 x + 5)$

Step 4.2

Apply the distributive property.

$f (2 x + 5) = 2 x (2 x) + 2 x \cdot 5 + 5 (2 x + 5)$

Step 4.3

Apply the distributive property.

$f (2 x + 5) = 2 x (2 x) + 2 x \cdot 5 + 5 (2 x) + 5 \cdot 5$

Step 5

Simplify and combine like terms.

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Step 5.1

Simplify each term.

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Step 5.1.1

Rewrite using the commutative property of multiplication.

$f (2 x + 5) = 2 \cdot (2 x \cdot x) + 2 x \cdot 5 + 5 (2 x) + 5 \cdot 5$

Step 5.1.2

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

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Step 5.1.2.1

Move $x$ .

$f (2 x + 5) = 2 \cdot (2 (x \cdot x)) + 2 x \cdot 5 + 5 (2 x) + 5 \cdot 5$

Step 5.1.2.2

Multiply $x$ by $x$ .

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

Step 5.1.3

Multiply $2$ by $2$ .

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

Step 5.1.4

Multiply $5$ by $2$ .

$f (2 x + 5) = 4 x^{2} + 10 x + 5 (2 x) + 5 \cdot 5$

Step 5.1.5

Multiply $2$ by $5$ .

$f (2 x + 5) = 4 x^{2} + 10 x + 10 x + 5 \cdot 5$

Step 5.1.6

Multiply $5$ by $5$ .

$f (2 x + 5) = 4 x^{2} + 10 x + 10 x + 25$

Step 5.2

Add $10 x$ and $10 x$ .

$f (2 x + 5) = 4 x^{2} + 20 x + 25$