Calculus Examples

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

Step 1

Write $- \frac{1}{3} x^{3} + x^{2}$ as a function.

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

Step 2

Find the $x$ values where the second derivative is equal to $0$ .

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Step 2.1

Find the second derivative.

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Step 2.1.1

Find the first derivative.

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Step 2.1.1.1

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

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

Step 2.1.1.2

Evaluate $\frac{d}{d x} [- \frac{1}{3} x^{3}]$ .

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Step 2.1.1.2.1

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

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

Step 2.1.1.2.2

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

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

Step 2.1.1.2.3

Multiply $3$ by $- 1$ .

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

Step 2.1.1.2.4

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

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

Step 2.1.1.2.5

Combine $\frac{- 3}{3}$ and $x^{2}$ .

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

Step 2.1.1.2.6

Cancel the common factor of $- 3$ and $3$ .

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Step 2.1.1.2.6.1

Factor $3$ out of $- 3 x^{2}$ .

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

Step 2.1.1.2.6.2

Cancel the common factors.

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Step 2.1.1.2.6.2.1

Factor $3$ out of $3$ .

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

Step 2.1.1.2.6.2.2

Cancel the common factor.

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

Step 2.1.1.2.6.2.3

Rewrite the expression.

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

Step 2.1.1.2.6.2.4

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

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

Step 2.1.1.3

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

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

Step 2.1.2

Find the second derivative.

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Step 2.1.2.1

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

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

Step 2.1.2.2

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

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Step 2.1.2.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

Step 2.1.2.2.2

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

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

Step 2.1.2.2.3

Multiply $2$ by $- 1$ .

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

Step 2.1.2.3

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

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Step 2.1.2.3.1

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

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

Step 2.1.2.3.2

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

$- 2 x + 2 \cdot 1$

Step 2.1.2.3.3

Multiply $2$ by $1$ .

$f'' (x) = - 2 x + 2$

Step 2.1.3

The second derivative of $f (x)$ with respect to $x$ is $- 2 x + 2$ .

$- 2 x + 2$

Step 2.2

Set the second derivative equal to $0$ then solve the equation $- 2 x + 2 = 0$ .

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Step 2.2.1

Set the second derivative equal to $0$ .

$- 2 x + 2 = 0$

Step 2.2.2

Subtract $2$ from both sides of the equation.

$- 2 x = - 2$

Step 2.2.3

Divide each term in $- 2 x = - 2$ by $- 2$ and simplify.

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Step 2.2.3.1

Divide each term in $- 2 x = - 2$ by $- 2$ .

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

Step 2.2.3.2

Simplify the left side.

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Step 2.2.3.2.1

Cancel the common factor of $- 2$ .

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Step 2.2.3.2.1.1

Cancel the common factor.

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

Step 2.2.3.2.1.2

Divide $x$ by $1$ .

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

Step 2.2.3.3

Simplify the right side.

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Step 2.2.3.3.1

Divide $- 2$ by $- 2$ .

$x = 1$

Step 3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, 1) \cup (1, \infty)$

Step 5

Substitute any number from the interval $(- \infty, 1)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

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

$f'' (0) = - 2 \cdot 0 + 2$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $- 2$ by $0$ .

$f'' (0) = 0 + 2$

Step 5.2.2

Add $0$ and $2$ .

$f'' (0) = 2$

Step 5.2.3

The final answer is $2$ .

$2$

Step 5.3

The graph is concave up on the interval $(- \infty, 1)$ because $f'' (0)$ is positive.

Concave up on $(- \infty, 1)$ since $f'' (x)$ is positive

Step 6

Substitute any number from the interval $(1, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

Replace the variable $x$ with $4$ in the expression.

$f'' (4) = - 2 \cdot 4 + 2$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $- 2$ by $4$ .

$f'' (4) = - 8 + 2$

Step 6.2.2

Add $- 8$ and $2$ .

$f'' (4) = - 6$

Step 6.2.3

The final answer is $- 6$ .

$- 6$

Step 6.3

The graph is concave down on the interval $(1, \infty)$ because $f'' (4)$ is negative.

Concave down on $(1, \infty)$ since $f'' (x)$ is negative

Step 7

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave up on $(- \infty, 1)$ since $f'' (x)$ is positive

Concave down on $(1, \infty)$ since $f'' (x)$ is negative

Step 8