Calculus Examples

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

Step 1

Write $y = - x^{3} + x - 5$ as a function.

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

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

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

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

Step 2.1.2

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

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Step 2.1.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{3}$ with respect to $x$ is $- \frac{d}{d x} [x^{3}]$ .

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

Step 2.1.2.2

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

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

Step 2.1.2.3

Multiply $3$ by $- 1$ .

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

Step 2.1.3

Differentiate.

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Step 2.1.3.1

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

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

Step 2.1.3.2

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

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

Step 2.1.3.3

Add $- 3 x^{2} + 1$ and $0$ .

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

Step 2.2

Find the second derivative.

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Step 2.2.1

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

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

Step 2.2.2

Evaluate $\frac{d}{d x} [- 3 x^{2}]$ .

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Step 2.2.2.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x^{2}$ with respect to $x$ is $- 3 \frac{d}{d x} [x^{2}]$ .

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

Step 2.2.2.2

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

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

Step 2.2.2.3

Multiply $2$ by $- 3$ .

$- 6 x + \frac{d}{d x} [1]$

Step 2.2.3

Differentiate using the Constant Rule.

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Step 2.2.3.1

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

$- 6 x + 0$

Step 2.2.3.2

Add $- 6 x$ and $0$ .

$f'' (x) = - 6 x$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $- 6 x$ .

$- 6 x$

Step 3

Set the second derivative equal to $0$ then solve the equation $- 6 x = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$- 6 x = 0$

Step 3.2

Divide each term in $- 6 x = 0$ by $- 6$ and simplify.

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Step 3.2.1

Divide each term in $- 6 x = 0$ by $- 6$ .

$\frac{- 6 x}{- 6} = \frac{0}{- 6}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $- 6$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{- 6 x}{- 6} = \frac{0}{- 6}$

Step 3.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{- 6}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Divide $0$ by $- 6$ .

$x = 0$

Step 4

Find the points where the second derivative is $0$ .

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Step 4.1

Substitute $0$ in $f (x) = - x^{3} + x - 5$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = - {(0)}^{3} + 0 - 5$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Remove parentheses.

$f (0) = - {(0)}^{3} + 0 - 5$

Step 4.1.2.2

Simplify each term.

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Step 4.1.2.2.1

Raising $0$ to any positive power yields $0$ .

$f (0) = - 0 + 0 - 5$

Step 4.1.2.2.2

Multiply $- 1$ by $0$ .

$f (0) = 0 + 0 - 5$

Step 4.1.2.3

Simplify by adding and subtracting.

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Step 4.1.2.3.1

Add $0$ and $0$ .

$f (0) = 0 - 5$

Step 4.1.2.3.2

Subtract $5$ from $0$ .

$f (0) = - 5$

Step 4.1.2.4

The final answer is $- 5$ .

$- 5$

Step 4.2

The point found by substituting $0$ in $f (x) = - x^{3} + x - 5$ is $(0, - 5)$ . This point can be an inflection point.

$(0, - 5)$

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

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

Step 6

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

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Step 6.1

Replace the variable $x$ with $- 0.1$ in the expression.

$f'' (- 0.1) = - 6 \cdot - 0.1$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $- 6$ by $- 0.1$ .

$f'' (- 0.1) = 0.6$

Step 6.2.2

The final answer is $0.6$ .

$0.6$

Step 6.3

At $- 0.1$ , the second derivative is $0.6$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, 0)$ .

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

Step 7

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

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Step 7.1

Replace the variable $x$ with $0.1$ in the expression.

$f'' (0.1) = - 6 \cdot 0.1$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $- 6$ by $0.1$ .

$f'' (0.1) = - 0.6$

Step 7.2.2

The final answer is $- 0.6$ .

$- 0.6$

Step 7.3

At $0.1$ , the second derivative is $- 0.6$ . Since this is negative, the second derivative is decreasing on the interval $(0, \infty)$

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

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(0, - 5)$ .

$(0, - 5)$

Step 9