Calculus Examples

$f (x) = 6 x^{2} - 4 x + 5$

Step 1

Find the derivative.

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Step 1.1

By the Sum Rule, the derivative of $6 x^{2} - 4 x + 5$ with respect to $x$ is $\frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [5]$

Step 1.2

Evaluate $\frac{d}{d x} [6 x^{2}]$ .

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Step 1.2.1

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{2}$ with respect to $x$ is $6 \frac{d}{d x} [x^{2}]$ .

$6 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [5]$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$6 (2 x) + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [5]$

Step 1.2.3

Multiply $2$ by $6$ .

$12 x + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [5]$

Step 1.3

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Step 1.3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$12 x - 4 \frac{d}{d x} [x] + \frac{d}{d x} [5]$

Step 1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$12 x - 4 \cdot 1 + \frac{d}{d x} [5]$

Step 1.3.3

Multiply $- 4$ by $1$ .

$12 x - 4 + \frac{d}{d x} [5]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$12 x - 4 + 0$

Step 1.4.2

Add $12 x - 4$ and $0$ .

$12 x - 4$

Step 2

Set the derivative equal to $0$ then solve the equation $12 x - 4 = 0$ .

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Step 2.1

Add $4$ to both sides of the equation.

$12 x = 4$

Step 2.2

Divide each term in $12 x = 4$ by $12$ and simplify.

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Step 2.2.1

Divide each term in $12 x = 4$ by $12$ .

$\frac{12 x}{12} = \frac{4}{12}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $12$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{12 x}{12} = \frac{4}{12}$

Step 2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{4}{12}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Cancel the common factor of $4$ and $12$ .

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Step 2.2.3.1.1

Factor $4$ out of $4$ .

$x = \frac{4 (1)}{12}$

Step 2.2.3.1.2

Cancel the common factors.

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Step 2.2.3.1.2.1

Factor $4$ out of $12$ .

$x = \frac{4 \cdot 1}{4 \cdot 3}$

Step 2.2.3.1.2.2

Cancel the common factor.

$x = \frac{4 \cdot 1}{4 \cdot 3}$

Step 2.2.3.1.2.3

Rewrite the expression.

$x = \frac{1}{3}$

Step 3

Solve the original function $f (x) = 6 x^{2} - 4 x + 5$ at $x = \frac{1}{3}$ .

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Step 3.1

Replace the variable $x$ with $\frac{1}{3}$ in the expression.

$f (\frac{1}{3}) = 6 {(\frac{1}{3})}^{2} - 4 (\frac{1}{3}) + 5$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Apply the product rule to $\frac{1}{3}$ .

$f (\frac{1}{3}) = 6 (\frac{1^{2}}{3^{2}}) - 4 (\frac{1}{3}) + 5$

Step 3.2.1.2

One to any power is one.

$f (\frac{1}{3}) = 6 (\frac{1}{3^{2}}) - 4 (\frac{1}{3}) + 5$

Step 3.2.1.3

Raise $3$ to the power of $2$ .

$f (\frac{1}{3}) = 6 (\frac{1}{9}) - 4 (\frac{1}{3}) + 5$

Step 3.2.1.4

Cancel the common factor of $3$ .

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Step 3.2.1.4.1

Factor $3$ out of $6$ .

$f (\frac{1}{3}) = 3 (2) (\frac{1}{9}) - 4 (\frac{1}{3}) + 5$

Step 3.2.1.4.2

Factor $3$ out of $9$ .

$f (\frac{1}{3}) = 3 \cdot (2 (\frac{1}{3 \cdot 3})) - 4 (\frac{1}{3}) + 5$

Step 3.2.1.4.3

Cancel the common factor.

$f (\frac{1}{3}) = 3 \cdot (2 (\frac{1}{3 \cdot 3})) - 4 (\frac{1}{3}) + 5$

Step 3.2.1.4.4

Rewrite the expression.

$f (\frac{1}{3}) = 2 (\frac{1}{3}) - 4 (\frac{1}{3}) + 5$

Step 3.2.1.5

Combine $2$ and $\frac{1}{3}$ .

$f (\frac{1}{3}) = \frac{2}{3} - 4 (\frac{1}{3}) + 5$

Step 3.2.1.6

Combine $- 4$ and $\frac{1}{3}$ .

$f (\frac{1}{3}) = \frac{2}{3} + \frac{- 4}{3} + 5$

Step 3.2.1.7

Move the negative in front of the fraction.

$f (\frac{1}{3}) = \frac{2}{3} - \frac{4}{3} + 5$

Step 3.2.2

Combine fractions.

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Step 3.2.2.1

Combine the numerators over the common denominator.

$f (\frac{1}{3}) = 5 + \frac{2 - 4}{3}$

Step 3.2.2.2

Simplify the expression.

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Step 3.2.2.2.1

Subtract $4$ from $2$ .

$f (\frac{1}{3}) = 5 + \frac{- 2}{3}$

Step 3.2.2.2.2

Move the negative in front of the fraction.

$f (\frac{1}{3}) = 5 - \frac{2}{3}$

Step 3.2.3

To write $5$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f (\frac{1}{3}) = 5 \cdot \frac{3}{3} - \frac{2}{3}$

Step 3.2.4

Combine $5$ and $\frac{3}{3}$ .

$f (\frac{1}{3}) = \frac{5 \cdot 3}{3} - \frac{2}{3}$

Step 3.2.5

Combine the numerators over the common denominator.

$f (\frac{1}{3}) = \frac{5 \cdot 3 - 2}{3}$

Step 3.2.6

Simplify the numerator.

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Step 3.2.6.1

Multiply $5$ by $3$ .

$f (\frac{1}{3}) = \frac{15 - 2}{3}$

Step 3.2.6.2

Subtract $2$ from $15$ .

$f (\frac{1}{3}) = \frac{13}{3}$

Step 3.2.7

The final answer is $\frac{13}{3}$ .

$\frac{13}{3}$

Step 4

The horizontal tangent line on function $f (x) = 6 x^{2} - 4 x + 5$ is $y = \frac{13}{3}$ .

$y = \frac{13}{3}$

Step 5