Calculus Examples

$f (x) = x^{4} - 50 x^{2} + 1$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate.

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Step 1.1.1.1

By the Sum Rule, the derivative of $x^{4} - 50 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 50 x^{2}] + \frac{d}{d x} [1]$ .

$f' (x) = \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 50 x^{2}) + \frac{d}{d x} (1)$

Step 1.1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f' (x) = 4 x^{3} + \frac{d}{d x} (- 50 x^{2}) + \frac{d}{d x} (1)$

Step 1.1.2

Evaluate $\frac{d}{d x} [- 50 x^{2}]$ .

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Step 1.1.2.1

Since $- 50$ is constant with respect to $x$ , the derivative of $- 50 x^{2}$ with respect to $x$ is $- 50 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 4 x^{3} - 50 \frac{d}{d x} x^{2} + \frac{d}{d x} (1)$

Step 1.1.2.2

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

$f' (x) = 4 x^{3} - 50 (2 x) + \frac{d}{d x} (1)$

Step 1.1.2.3

Multiply $2$ by $- 50$ .

$f' (x) = 4 x^{3} - 100 x + \frac{d}{d x} (1)$

Step 1.1.3

Differentiate using the Constant Rule.

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Step 1.1.3.1

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

$f' (x) = 4 x^{3} - 100 x + 0$

Step 1.1.3.2

Add $4 x^{3} - 100 x$ and $0$ .

$f' (x) = 4 x^{3} - 100 x$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $4 x^{3} - 100 x$ .

$4 x^{3} - 100 x$

Step 2

Set the first derivative equal to $0$ then solve the equation $4 x^{3} - 100 x = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$4 x^{3} - 100 x = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $4 x$ out of $4 x^{3} - 100 x$ .

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Step 2.2.1.1

Factor $4 x$ out of $4 x^{3}$ .

$4 x (x^{2}) - 100 x = 0$

Step 2.2.1.2

Factor $4 x$ out of $- 100 x$ .

$4 x (x^{2}) + 4 x (- 25) = 0$

Step 2.2.1.3

Factor $4 x$ out of $4 x (x^{2}) + 4 x (- 25)$ .

$4 x (x^{2} - 25) = 0$

Step 2.2.2

Rewrite $25$ as $5^{2}$ .

$4 x (x^{2} - 5^{2}) = 0$

Step 2.2.3

Factor.

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Step 2.2.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

$4 x ((x + 5) (x - 5)) = 0$

Step 2.2.3.2

Remove unnecessary parentheses.

$4 x (x + 5) (x - 5) = 0$

Step 2.3

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

$x = 0$

$x + 5 = 0$

$x - 5 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 2.5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.6

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 2.6.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.7

The final solution is all the values that make $4 x (x + 5) (x - 5) = 0$ true.

$x = 0, - 5, 5$

Step 3

The values which make the derivative equal to $0$ are $0, - 5, 5$ .

$0, - 5, 5$

Step 4

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the derivative $0$ or undefined.

$(- \infty, - 5) \cup (- 5, 0) \cup (0, 5) \cup (5, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 5)$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 6$ in the expression.

$f' (- 6) = 4 {(- 6)}^{3} - 100 \cdot - 6$

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 $- 6$ to the power of $3$ .

$f' (- 6) = 4 \cdot - 216 - 100 \cdot - 6$

Step 5.2.1.2

Multiply $4$ by $- 216$ .

$f' (- 6) = - 864 - 100 \cdot - 6$

Step 5.2.1.3

Multiply $- 100$ by $- 6$ .

$f' (- 6) = - 864 + 600$

Step 5.2.2

Add $- 864$ and $600$ .

$f' (- 6) = - 264$

Step 5.2.3

The final answer is $- 264$ .

$- 264$

Step 5.3

At $x = - 6$ the derivative is $- 264$ . Since this is negative, the function is decreasing on $(- \infty, - 5)$ .

Decreasing on $(- \infty, - 5)$ since $f' (x) < 0$

Step 6

Substitute a value from the interval $(- 5, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- \frac{5}{2}$ in the expression.

$f' (- \frac{5}{2}) = 4 {(- \frac{5}{2})}^{3} - 100 (- \frac{5}{2})$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 6.2.1.1.1

Apply the product rule to $- \frac{5}{2}$ .

$f' (- \frac{5}{2}) = 4 ({(- 1)}^{3} {(\frac{5}{2})}^{3}) - 100 (- \frac{5}{2})$

Step 6.2.1.1.2

Apply the product rule to $\frac{5}{2}$ .

$f' (- \frac{5}{2}) = 4 ({(- 1)}^{3} (\frac{5^{3}}{2^{3}})) - 100 (- \frac{5}{2})$

Step 6.2.1.2

Raise $- 1$ to the power of $3$ .

$f' (- \frac{5}{2}) = 4 (- \frac{5^{3}}{2^{3}}) - 100 (- \frac{5}{2})$

Step 6.2.1.3

Raise $5$ to the power of $3$ .

$f' (- \frac{5}{2}) = 4 (- \frac{125}{2^{3}}) - 100 (- \frac{5}{2})$

Step 6.2.1.4

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

$f' (- \frac{5}{2}) = 4 (- \frac{125}{8}) - 100 (- \frac{5}{2})$

Step 6.2.1.5

Cancel the common factor of $4$ .

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Step 6.2.1.5.1

Move the leading negative in $- \frac{125}{8}$ into the numerator.

$f' (- \frac{5}{2}) = 4 (\frac{- 125}{8}) - 100 (- \frac{5}{2})$

Step 6.2.1.5.2

Factor $4$ out of $8$ .

$f' (- \frac{5}{2}) = 4 (\frac{- 125}{4 (2)}) - 100 (- \frac{5}{2})$

Step 6.2.1.5.3

Cancel the common factor.

$f' (- \frac{5}{2}) = 4 (\frac{- 125}{4 \cdot 2}) - 100 (- \frac{5}{2})$

Step 6.2.1.5.4

Rewrite the expression.

$f' (- \frac{5}{2}) = \frac{- 125}{2} - 100 (- \frac{5}{2})$

Step 6.2.1.6

Move the negative in front of the fraction.

$f' (- \frac{5}{2}) = - \frac{125}{2} - 100 (- \frac{5}{2})$

Step 6.2.1.7

Cancel the common factor of $2$ .

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Step 6.2.1.7.1

Move the leading negative in $- \frac{5}{2}$ into the numerator.

$f' (- \frac{5}{2}) = - \frac{125}{2} - 100 (\frac{- 5}{2})$

Step 6.2.1.7.2

Factor $2$ out of $- 100$ .

$f' (- \frac{5}{2}) = - \frac{125}{2} + 2 (- 50) (\frac{- 5}{2})$

Step 6.2.1.7.3

Cancel the common factor.

$f' (- \frac{5}{2}) = - \frac{125}{2} + 2 \cdot (- 50 (\frac{- 5}{2}))$

Step 6.2.1.7.4

Rewrite the expression.

$f' (- \frac{5}{2}) = - \frac{125}{2} - 50 \cdot - 5$

Step 6.2.1.8

Multiply $- 50$ by $- 5$ .

$f' (- \frac{5}{2}) = - \frac{125}{2} + 250$

Step 6.2.2

To write $250$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (- \frac{5}{2}) = - \frac{125}{2} + 250 \cdot \frac{2}{2}$

Step 6.2.3

Combine $250$ and $\frac{2}{2}$ .

$f' (- \frac{5}{2}) = - \frac{125}{2} + \frac{250 \cdot 2}{2}$

Step 6.2.4

Combine the numerators over the common denominator.

$f' (- \frac{5}{2}) = \frac{- 125 + 250 \cdot 2}{2}$

Step 6.2.5

Simplify the numerator.

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Step 6.2.5.1

Multiply $250$ by $2$ .

$f' (- \frac{5}{2}) = \frac{- 125 + 500}{2}$

Step 6.2.5.2

Add $- 125$ and $500$ .

$f' (- \frac{5}{2}) = \frac{375}{2}$

Step 6.2.6

The final answer is $\frac{375}{2}$ .

$\frac{375}{2}$

Step 6.3

At $x = - \frac{5}{2}$ the derivative is $\frac{375}{2}$ . Since this is positive, the function is increasing on $(- 5, 0)$ .

Increasing on $(- 5, 0)$ since $f' (x) > 0$

Step 7

Substitute a value from the interval $(0, 5)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $\frac{5}{2}$ in the expression.

$f' (\frac{5}{2}) = 4 {(\frac{5}{2})}^{3} - 100 (\frac{5}{2})$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Apply the product rule to $\frac{5}{2}$ .

$f' (\frac{5}{2}) = 4 (\frac{5^{3}}{2^{3}}) - 100 (\frac{5}{2})$

Step 7.2.1.2

Raise $5$ to the power of $3$ .

$f' (\frac{5}{2}) = 4 (\frac{125}{2^{3}}) - 100 (\frac{5}{2})$

Step 7.2.1.3

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

$f' (\frac{5}{2}) = 4 (\frac{125}{8}) - 100 (\frac{5}{2})$

Step 7.2.1.4

Cancel the common factor of $4$ .

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Step 7.2.1.4.1

Factor $4$ out of $8$ .

$f' (\frac{5}{2}) = 4 (\frac{125}{4 (2)}) - 100 (\frac{5}{2})$

Step 7.2.1.4.2

Cancel the common factor.

$f' (\frac{5}{2}) = 4 (\frac{125}{4 \cdot 2}) - 100 (\frac{5}{2})$

Step 7.2.1.4.3

Rewrite the expression.

$f' (\frac{5}{2}) = \frac{125}{2} - 100 (\frac{5}{2})$

Step 7.2.1.5

Cancel the common factor of $2$ .

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Step 7.2.1.5.1

Factor $2$ out of $- 100$ .

$f' (\frac{5}{2}) = \frac{125}{2} + 2 (- 50) (\frac{5}{2})$

Step 7.2.1.5.2

Cancel the common factor.

$f' (\frac{5}{2}) = \frac{125}{2} + 2 \cdot (- 50 (\frac{5}{2}))$

Step 7.2.1.5.3

Rewrite the expression.

$f' (\frac{5}{2}) = \frac{125}{2} - 50 \cdot 5$

Step 7.2.1.6

Multiply $- 50$ by $5$ .

$f' (\frac{5}{2}) = \frac{125}{2} - 250$

Step 7.2.2

To write $- 250$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{125}{2} - 250 \cdot \frac{2}{2}$

Step 7.2.3

Combine $- 250$ and $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{125}{2} + \frac{- 250 \cdot 2}{2}$

Step 7.2.4

Combine the numerators over the common denominator.

$f' (\frac{5}{2}) = \frac{125 - 250 \cdot 2}{2}$

Step 7.2.5

Simplify the numerator.

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Step 7.2.5.1

Multiply $- 250$ by $2$ .

$f' (\frac{5}{2}) = \frac{125 - 500}{2}$

Step 7.2.5.2

Subtract $500$ from $125$ .

$f' (\frac{5}{2}) = \frac{- 375}{2}$

Step 7.2.6

Move the negative in front of the fraction.

$f' (\frac{5}{2}) = - \frac{375}{2}$

Step 7.2.7

The final answer is $- \frac{375}{2}$ .

$- \frac{375}{2}$

Step 7.3

At $x = \frac{5}{2}$ the derivative is $- \frac{375}{2}$ . Since this is negative, the function is decreasing on $(0, 5)$ .

Decreasing on $(0, 5)$ since $f' (x) < 0$

Step 8

Substitute a value from the interval $(5, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $6$ in the expression.

$f' (6) = 4 {(6)}^{3} - 100 \cdot 6$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

Raise $6$ to the power of $3$ .

$f' (6) = 4 \cdot 216 - 100 \cdot 6$

Step 8.2.1.2

Multiply $4$ by $216$ .

$f' (6) = 864 - 100 \cdot 6$

Step 8.2.1.3

Multiply $- 100$ by $6$ .

$f' (6) = 864 - 600$

Step 8.2.2

Subtract $600$ from $864$ .

$f' (6) = 264$

Step 8.2.3

The final answer is $264$ .

$264$

Step 8.3

At $x = 6$ the derivative is $264$ . Since this is positive, the function is increasing on $(5, \infty)$ .

Increasing on $(5, \infty)$ since $f' (x) > 0$

Step 9

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- 5, 0), (5, \infty)$

Decreasing on: $(- \infty, - 5), (0, 5)$

Step 10