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Calculus Examples

f (x) = 6 x^{2} + x

$f (x) = 6 x^{2} + x$ ,

g (x) = x + 1

$g (x) = x + 1$ ,

f (x) - g (x)

$f (x) - g (x)$

Step 1

Replace the function designators in

f (x) - g x

$f (x) - g x$ with the actual functions.

(6 x^{2} + x) - (x + 1)

$(6 x^{2} + x) - (x + 1)$

Step 2

Simplify

(6 x^{2} + x) - (x + 1)

$(6 x^{2} + x) - (x + 1)$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Apply the distributive property.

6 x^{2} + x - x - 1 \cdot 1

$6 x^{2} + x - x - 1 \cdot 1$

Step 2.1.2

Multiply

- 1

$- 1$ by

1

$1$ .

6 x^{2} + x - x - 1

$6 x^{2} + x - x - 1$

6 x^{2} + x - x - 1

$6 x^{2} + x - x - 1$

Step 2.2

Combine the opposite terms in

6 x^{2} + x - x - 1

$6 x^{2} + x - x - 1$ .

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Step 2.2.1

Subtract

x

$x$ from

x

$x$ .

6 x^{2} + 0 - 1

$6 x^{2} + 0 - 1$

Step 2.2.2

Add

6 x^{2}

$6 x^{2}$ and

0

$0$ .

6 x^{2} - 1

$6 x^{2} - 1$

6 x^{2} - 1

$6 x^{2} - 1$

6 x^{2} - 1

$6 x^{2} - 1$

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⎡ ⎢ ⎣ \begin{matrix} x^{2} & \frac{1}{2} \sqrt{π} & \int x d x \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$