Calculus Examples

$y = x^{3} - 5 x^{2} + 3 x + 1$

Step 1

Set $y$ as a function of $x$ .

$f (x) = x^{3} - 5 x^{2} + 3 x + 1$

Step 2

Find the derivative.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $x^{3} - 5 x^{2} + 3 x + 1$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 5 x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 5 x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]$

Step 2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$3 x^{2} + \frac{d}{d x} [- 5 x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]$

Step 2.2

Evaluate $\frac{d}{d x} [- 5 x^{2}]$ .

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Step 2.2.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x^{2}$ with respect to $x$ is $- 5 \frac{d}{d x} [x^{2}]$ .

$3 x^{2} - 5 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]$

Step 2.2.2

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

$3 x^{2} - 5 (2 x) + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]$

Step 2.2.3

Multiply $2$ by $- 5$ .

$3 x^{2} - 10 x + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]$

Step 2.3

Evaluate $\frac{d}{d x} [3 x]$ .

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Step 2.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$3 x^{2} - 10 x + 3 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 2.3.2

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

$3 x^{2} - 10 x + 3 \cdot 1 + \frac{d}{d x} [1]$

Step 2.3.3

Multiply $3$ by $1$ .

$3 x^{2} - 10 x + 3 + \frac{d}{d x} [1]$

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$3 x^{2} - 10 x + 3 + 0$

Step 2.4.2

Add $3 x^{2} - 10 x + 3$ and $0$ .

$3 x^{2} - 10 x + 3$

Step 3

Set the derivative equal to $0$ then solve the equation $3 x^{2} - 10 x + 3 = 0$ .

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Step 3.1

Factor by grouping.

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Step 3.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot 3 = 9$ and whose sum is $b = - 10$ .

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Step 3.1.1.1

Factor $- 10$ out of $- 10 x$ .

$3 x^{2} - 10 x + 3 = 0$

Step 3.1.1.2

Rewrite $- 10$ as $- 1$ plus $- 9$

$3 x^{2} + (- 1 - 9) x + 3 = 0$

Step 3.1.1.3

Apply the distributive property.

$3 x^{2} - 1 x - 9 x + 3 = 0$

Step 3.1.2

Factor out the greatest common factor from each group.

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Step 3.1.2.1

Group the first two terms and the last two terms.

$(3 x^{2} - 1 x) - 9 x + 3 = 0$

Step 3.1.2.2

Factor out the greatest common factor (GCF) from each group.

$x (3 x - 1) - 3 (3 x - 1) = 0$

Step 3.1.3

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$(3 x - 1) (x - 3) = 0$

Step 3.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 x - 1 = 0$

$x - 3 = 0$

Step 3.3

Set $3 x - 1$ equal to $0$ and solve for $x$ .

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Step 3.3.1

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Step 3.3.2

Solve $3 x - 1 = 0$ for $x$ .

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Step 3.3.2.1

Add $1$ to both sides of the equation.

$3 x = 1$

Step 3.3.2.2

Divide each term in $3 x = 1$ by $3$ and simplify.

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Step 3.3.2.2.1

Divide each term in $3 x = 1$ by $3$ .

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

Step 3.3.2.2.2

Simplify the left side.

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Step 3.3.2.2.2.1

Cancel the common factor of $3$ .

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Step 3.3.2.2.2.1.1

Cancel the common factor.

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

Step 3.3.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 3.4

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 3.4.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 3.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 3.5

The final solution is all the values that make $(3 x - 1) (x - 3) = 0$ true.

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

Step 4

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

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Step 4.1

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

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

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

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

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

Step 4.2.1.2

One to any power is one.

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

Step 4.2.1.3

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

$f (\frac{1}{3}) = \frac{1}{27} - 5 {(\frac{1}{3})}^{2} + 3 (\frac{1}{3}) + 1$

Step 4.2.1.4

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

$f (\frac{1}{3}) = \frac{1}{27} - 5 \frac{1^{2}}{3^{2}} + 3 (\frac{1}{3}) + 1$

Step 4.2.1.5

One to any power is one.

$f (\frac{1}{3}) = \frac{1}{27} - 5 \frac{1}{3^{2}} + 3 (\frac{1}{3}) + 1$

Step 4.2.1.6

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

$f (\frac{1}{3}) = \frac{1}{27} - 5 (\frac{1}{9}) + 3 (\frac{1}{3}) + 1$

Step 4.2.1.7

Combine $- 5$ and $\frac{1}{9}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{- 5}{9} + 3 (\frac{1}{3}) + 1$

Step 4.2.1.8

Move the negative in front of the fraction.

$f (\frac{1}{3}) = \frac{1}{27} - \frac{5}{9} + 3 (\frac{1}{3}) + 1$

Step 4.2.1.9

Cancel the common factor of $3$ .

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Step 4.2.1.9.1

Cancel the common factor.

$f (\frac{1}{3}) = \frac{1}{27} - \frac{5}{9} + 3 (\frac{1}{3}) + 1$

Step 4.2.1.9.2

Rewrite the expression.

$f (\frac{1}{3}) = \frac{1}{27} - \frac{5}{9} + 1 + 1$

Step 4.2.2

Find the common denominator.

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Step 4.2.2.1

Multiply $\frac{5}{9}$ by $\frac{3}{3}$ .

$f (\frac{1}{3}) = \frac{1}{27} - (\frac{5}{9} \cdot \frac{3}{3}) + 1 + 1$

Step 4.2.2.2

Multiply $\frac{5}{9}$ by $\frac{3}{3}$ .

$f (\frac{1}{3}) = \frac{1}{27} - \frac{5 \cdot 3}{9 \cdot 3} + 1 + 1$

Step 4.2.2.3

Write $1$ as a fraction with denominator $1$ .

$f (\frac{1}{3}) = \frac{1}{27} - \frac{5 \cdot 3}{9 \cdot 3} + \frac{1}{1} + 1$

Step 4.2.2.4

Multiply $\frac{1}{1}$ by $\frac{27}{27}$ .

$f (\frac{1}{3}) = \frac{1}{27} - \frac{5 \cdot 3}{9 \cdot 3} + \frac{1}{1} \cdot \frac{27}{27} + 1$

Step 4.2.2.5

Multiply $\frac{1}{1}$ by $\frac{27}{27}$ .

$f (\frac{1}{3}) = \frac{1}{27} - \frac{5 \cdot 3}{9 \cdot 3} + \frac{27}{27} + 1$

Step 4.2.2.6

Write $1$ as a fraction with denominator $1$ .

$f (\frac{1}{3}) = \frac{1}{27} - \frac{5 \cdot 3}{9 \cdot 3} + \frac{27}{27} + \frac{1}{1}$

Step 4.2.2.7

Multiply $\frac{1}{1}$ by $\frac{27}{27}$ .

$f (\frac{1}{3}) = \frac{1}{27} - \frac{5 \cdot 3}{9 \cdot 3} + \frac{27}{27} + \frac{1}{1} \cdot \frac{27}{27}$

Step 4.2.2.8

Multiply $\frac{1}{1}$ by $\frac{27}{27}$ .

$f (\frac{1}{3}) = \frac{1}{27} - \frac{5 \cdot 3}{9 \cdot 3} + \frac{27}{27} + \frac{27}{27}$

Step 4.2.2.9

Reorder the factors of $9 \cdot 3$ .

$f (\frac{1}{3}) = \frac{1}{27} - \frac{5 \cdot 3}{3 \cdot 9} + \frac{27}{27} + \frac{27}{27}$

Step 4.2.2.10

Multiply $3$ by $9$ .

$f (\frac{1}{3}) = \frac{1}{27} - \frac{5 \cdot 3}{27} + \frac{27}{27} + \frac{27}{27}$

Step 4.2.3

Combine the numerators over the common denominator.

$f (\frac{1}{3}) = \frac{1 - 5 \cdot 3 + 27 + 27}{27}$

Step 4.2.4

Simplify the expression.

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Step 4.2.4.1

Multiply $- 5$ by $3$ .

$f (\frac{1}{3}) = \frac{1 - 15 + 27 + 27}{27}$

Step 4.2.4.2

Subtract $15$ from $1$ .

$f (\frac{1}{3}) = \frac{- 14 + 27 + 27}{27}$

Step 4.2.4.3

Add $- 14$ and $27$ .

$f (\frac{1}{3}) = \frac{13 + 27}{27}$

Step 4.2.4.4

Add $13$ and $27$ .

$f (\frac{1}{3}) = \frac{40}{27}$

Step 4.2.5

The final answer is $\frac{40}{27}$ .

$\frac{40}{27}$

Step 5

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

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Step 5.1

Replace the variable $x$ with $3$ in the expression.

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

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

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

$f (3) = 27 - 5 {(3)}^{2} + 3 (3) + 1$

Step 5.2.1.2

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

$f (3) = 27 - 5 \cdot 9 + 3 (3) + 1$

Step 5.2.1.3

Multiply $- 5$ by $9$ .

$f (3) = 27 - 45 + 3 (3) + 1$

Step 5.2.1.4

Multiply $3$ by $3$ .

$f (3) = 27 - 45 + 9 + 1$

Step 5.2.2

Simplify by adding and subtracting.

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Step 5.2.2.1

Subtract $45$ from $27$ .

$f (3) = - 18 + 9 + 1$

Step 5.2.2.2

Add $- 18$ and $9$ .

$f (3) = - 9 + 1$

Step 5.2.2.3

Add $- 9$ and $1$ .

$f (3) = - 8$

Step 5.2.3

The final answer is $- 8$ .

$- 8$

Step 6

The horizontal tangent lines on function $f (x) = x^{3} - 5 x^{2} + 3 x + 1$ are $y = \frac{40}{27}, y = - 8$ .

$y = \frac{40}{27}, y = - 8$

Step 7