Calculus Examples

$f (x) = - 5 x^{4} - 4 x^{2} + 9$

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

By the Sum Rule, the derivative of $- 5 x^{4} - 4 x^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [- 5 x^{4}] + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [9]$ .

$\frac{d}{d x} [- 5 x^{4}] + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [9]$

Step 1.1.2

Evaluate $\frac{d}{d x} [- 5 x^{4}]$ .

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Step 1.1.2.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x^{4}$ with respect to $x$ is $- 5 \frac{d}{d x} [x^{4}]$ .

$- 5 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [9]$

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 $n = 4$ .

$- 5 (4 x^{3}) + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [9]$

Step 1.1.2.3

Multiply $4$ by $- 5$ .

$- 20 x^{3} + \frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [9]$

Step 1.1.3

Evaluate $\frac{d}{d x} [- 4 x^{2}]$ .

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Step 1.1.3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x^{2}$ with respect to $x$ is $- 4 \frac{d}{d x} [x^{2}]$ .

$- 20 x^{3} - 4 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [9]$

Step 1.1.3.2

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

$- 20 x^{3} - 4 (2 x) + \frac{d}{d x} [9]$

Step 1.1.3.3

Multiply $2$ by $- 4$ .

$- 20 x^{3} - 8 x + \frac{d}{d x} [9]$

Step 1.1.4

Differentiate using the Constant Rule.

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Step 1.1.4.1

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$- 20 x^{3} - 8 x + 0$

Step 1.1.4.2

Add $- 20 x^{3} - 8 x$ and $0$ .

$f' (x) = - 20 x^{3} - 8 x$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- 20 x^{3} - 8 x$ .

$- 20 x^{3} - 8 x$

Step 2

Set the first derivative equal to $0$ then solve the equation $- 20 x^{3} - 8 x = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$- 20 x^{3} - 8 x = 0$

Step 2.2

Factor $- 4 x$ out of $- 20 x^{3} - 8 x$ .

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Step 2.2.1

Factor $- 4 x$ out of $- 20 x^{3}$ .

$- 4 x (5 x^{2}) - 8 x = 0$

Step 2.2.2

Factor $- 4 x$ out of $- 8 x$ .

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

Step 2.2.3

Factor $- 4 x$ out of $- 4 x (5 x^{2}) - 4 x (2)$ .

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

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 $0$ .

$x = 0$

$5 x^{2} + 2 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $5 x^{2} + 2$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $5 x^{2} + 2$ equal to $0$ .

$5 x^{2} + 2 = 0$

Step 2.5.2

Solve $5 x^{2} + 2 = 0$ for $x$ .

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Step 2.5.2.1

Subtract $2$ from both sides of the equation.

$5 x^{2} = - 2$

Step 2.5.2.2

Divide each term in $5 x^{2} = - 2$ by $5$ and simplify.

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Step 2.5.2.2.1

Divide each term in $5 x^{2} = - 2$ by $5$ .

$\frac{5 x^{2}}{5} = \frac{- 2}{5}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $5$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

$\frac{5 x^{2}}{5} = \frac{- 2}{5}$

Step 2.5.2.2.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 2}{5}$

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Move the negative in front of the fraction.

$x^{2} = - \frac{2}{5}$

Step 2.5.2.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- \frac{2}{5}}$

Step 2.5.2.4

Simplify $\pm \sqrt{- \frac{2}{5}}$ .

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Step 2.5.2.4.1

Rewrite $- 1$ as $i^{2}$ .

$x = \pm \sqrt{i^{2} \frac{2}{5}}$

Step 2.5.2.4.2

Pull terms out from under the radical.

$x = \pm i \sqrt{\frac{2}{5}}$

Step 2.5.2.4.3

Rewrite $\sqrt{\frac{2}{5}}$ as $\frac{\sqrt{2}}{\sqrt{5}}$ .

$x = \pm i \frac{\sqrt{2}}{\sqrt{5}}$

Step 2.5.2.4.4

Multiply $\frac{\sqrt{2}}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

$x = \pm i (\frac{\sqrt{2}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}})$

Step 2.5.2.4.5

Combine and simplify the denominator.

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Step 2.5.2.4.5.1

Multiply $\frac{\sqrt{2}}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

$x = \pm i \frac{\sqrt{2} \sqrt{5}}{\sqrt{5} \sqrt{5}}$

Step 2.5.2.4.5.2

Raise $\sqrt{5}$ to the power of $1$ .

$x = \pm i \frac{\sqrt{2} \sqrt{5}}{{\sqrt{5}}^{1} \sqrt{5}}$

Step 2.5.2.4.5.3

Raise $\sqrt{5}$ to the power of $1$ .

$x = \pm i \frac{\sqrt{2} \sqrt{5}}{{\sqrt{5}}^{1} {\sqrt{5}}^{1}}$

Step 2.5.2.4.5.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x = \pm i \frac{\sqrt{2} \sqrt{5}}{{\sqrt{5}}^{1 + 1}}$

Step 2.5.2.4.5.5

Add $1$ and $1$ .

$x = \pm i \frac{\sqrt{2} \sqrt{5}}{{\sqrt{5}}^{2}}$

Step 2.5.2.4.5.6

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

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Step 2.5.2.4.5.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$x = \pm i \frac{\sqrt{2} \sqrt{5}}{{(5^{\frac{1}{2}})}^{2}}$

Step 2.5.2.4.5.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x = \pm i \frac{\sqrt{2} \sqrt{5}}{5^{\frac{1}{2} \cdot 2}}$

Step 2.5.2.4.5.6.3

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

$x = \pm i \frac{\sqrt{2} \sqrt{5}}{5^{\frac{2}{2}}}$

Step 2.5.2.4.5.6.4

Cancel the common factor of $2$ .

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Step 2.5.2.4.5.6.4.1

Cancel the common factor.

$x = \pm i \frac{\sqrt{2} \sqrt{5}}{5^{\frac{2}{2}}}$

Step 2.5.2.4.5.6.4.2

Rewrite the expression.

$x = \pm i \frac{\sqrt{2} \sqrt{5}}{5^{1}}$

Step 2.5.2.4.5.6.5

Evaluate the exponent.

$x = \pm i \frac{\sqrt{2} \sqrt{5}}{5}$

Step 2.5.2.4.6

Simplify the numerator.

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Step 2.5.2.4.6.1

Combine using the product rule for radicals.

$x = \pm i \frac{\sqrt{2 \cdot 5}}{5}$

Step 2.5.2.4.6.2

Multiply $2$ by $5$ .

$x = \pm i \frac{\sqrt{10}}{5}$

Step 2.5.2.4.7

Combine $i$ and $\frac{\sqrt{10}}{5}$ .

$x = \pm \frac{i \sqrt{10}}{5}$

Step 2.5.2.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.5.2.5.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \frac{i \sqrt{10}}{5}$

Step 2.5.2.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \frac{i \sqrt{10}}{5}$

Step 2.5.2.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{i \sqrt{10}}{5}, - \frac{i \sqrt{10}}{5}$

Step 2.6

The final solution is all the values that make $- 4 x (5 x^{2} + 2) = 0$ true.

$x = 0, \frac{i \sqrt{10}}{5}, - \frac{i \sqrt{10}}{5}$

Step 3

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

$0$

Step 4

After finding the point that makes the derivative $f' (x) = - 20 x^{3} - 8 x$ equal to $0$ or undefined, the interval to check where $f (x) = - 5 x^{4} - 4 x^{2} + 9$ is increasing and where it is decreasing is $(- \infty, 0) \cup (0, \infty)$ .

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

Step 5

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

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Step 5.1

Replace the variable $x$ with $- 1$ in the expression.

$f' (- 1) = - 20 {(- 1)}^{3} - 8 \cdot - 1$

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

$f' (- 1) = - 20 \cdot - 1 - 8 \cdot - 1$

Step 5.2.1.2

Multiply $- 20$ by $- 1$ .

$f' (- 1) = 20 - 8 \cdot - 1$

Step 5.2.1.3

Multiply $- 8$ by $- 1$ .

$f' (- 1) = 20 + 8$

Step 5.2.2

Add $20$ and $8$ .

$f' (- 1) = 28$

Step 5.2.3

The final answer is $28$ .

$28$

Step 5.3

At $x = - 1$ the derivative is $28$ . Since this is positive, the function is increasing on $(- \infty, 0)$ .

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

Step 6

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

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Step 6.1

Replace the variable $x$ with $1$ in the expression.

$f' (1) = - 20 {(1)}^{3} - 8 \cdot 1$

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

One to any power is one.

$f' (1) = - 20 \cdot 1 - 8 \cdot 1$

Step 6.2.1.2

Multiply $- 20$ by $1$ .

$f' (1) = - 20 - 8 \cdot 1$

Step 6.2.1.3

Multiply $- 8$ by $1$ .

$f' (1) = - 20 - 8$

Step 6.2.2

Subtract $8$ from $- 20$ .

$f' (1) = - 28$

Step 6.2.3

The final answer is $- 28$ .

$- 28$

Step 6.3

At $x = 1$ the derivative is $- 28$ . Since this is negative, the function is decreasing on $(0, \infty)$ .

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

Step 7

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

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

Decreasing on: $(0, \infty)$

Step 8