Calculus Examples

$f (x) = 7 x + 5 x^{- 1}$

Step 1

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

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Step 1.1

Find the second derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

By the Sum Rule, the derivative of $7 x + 5 x^{- 1}$ with respect to $x$ is $\frac{d}{d x} [7 x] + \frac{d}{d x} [5 x^{- 1}]$ .

$\frac{d}{d x} [7 x] + \frac{d}{d x} [5 x^{- 1}]$

Step 1.1.1.2

Evaluate $\frac{d}{d x} [7 x]$ .

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Step 1.1.1.2.1

Since $7$ is constant with respect to $x$ , the derivative of $7 x$ with respect to $x$ is $7 \frac{d}{d x} [x]$ .

$7 \frac{d}{d x} [x] + \frac{d}{d x} [5 x^{- 1}]$

Step 1.1.1.2.2

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

$7 \cdot 1 + \frac{d}{d x} [5 x^{- 1}]$

Step 1.1.1.2.3

Multiply $7$ by $1$ .

$7 + \frac{d}{d x} [5 x^{- 1}]$

Step 1.1.1.3

Evaluate $\frac{d}{d x} [5 x^{- 1}]$ .

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Step 1.1.1.3.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{- 1}$ with respect to $x$ is $5 \frac{d}{d x} [x^{- 1}]$ .

$7 + 5 \frac{d}{d x} [x^{- 1}]$

Step 1.1.1.3.2

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

$7 + 5 (- x^{- 2})$

Step 1.1.1.3.3

Multiply $- 1$ by $5$ .

$7 - 5 x^{- 2}$

Step 1.1.1.4

Reorder terms.

$f' (x) = - 5 x^{- 2} + 7$

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

By the Sum Rule, the derivative of $- 5 x^{- 2} + 7$ with respect to $x$ is $\frac{d}{d x} [- 5 x^{- 2}] + \frac{d}{d x} [7]$ .

$\frac{d}{d x} [- 5 x^{- 2}] + \frac{d}{d x} [7]$

Step 1.1.2.2

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

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Step 1.1.2.2.1

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

$- 5 \frac{d}{d x} [x^{- 2}] + \frac{d}{d x} [7]$

Step 1.1.2.2.2

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

$- 5 (- 2 x^{- 3}) + \frac{d}{d x} [7]$

Step 1.1.2.2.3

Multiply $- 2$ by $- 5$ .

$10 x^{- 3} + \frac{d}{d x} [7]$

Step 1.1.2.3

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$10 x^{- 3} + 0$

Step 1.1.2.4

Simplify.

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Step 1.1.2.4.1

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

$10 x^{- 3}$

Step 1.1.2.4.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$10 \frac{1}{x^{3}}$

Step 1.1.2.4.3

Combine $10$ and $\frac{1}{x^{3}}$ .

$f'' (x) = \frac{10}{x^{3}}$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{10}{x^{3}}$ .

$\frac{10}{x^{3}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $\frac{10}{x^{3}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$\frac{10}{x^{3}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$10 = 0$

Step 1.2.3

Since $10 \neq 0$ , there are no solutions.

No solution

Step 2

Find the domain of $f (x) = 7 x + 5 x^{- 1}$ .

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Step 2.1

Set the base in $x^{- 1}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 2.2

The domain is all values of $x$ that make the expression defined.

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 0}$

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 0}$

Step 3

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

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

Step 4

Substitute any number from the interval $(- \infty, 0)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

Replace the variable $x$ with $- 2$ in the expression.

$f'' (- 2) = \frac{10}{{(- 2)}^{3}}$

Step 4.2

Simplify the result.

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Step 4.2.1

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

$f'' (- 2) = \frac{10}{- 8}$

Step 4.2.2

Cancel the common factor of $10$ and $- 8$ .

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Step 4.2.2.1

Factor $2$ out of $10$ .

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

Step 4.2.2.2

Cancel the common factors.

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Step 4.2.2.2.1

Factor $2$ out of $- 8$ .

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

Step 4.2.2.2.2

Cancel the common factor.

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

Step 4.2.2.2.3

Rewrite the expression.

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

Step 4.2.3

Move the negative in front of the fraction.

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

Step 4.2.4

The final answer is $- \frac{5}{4}$ .

$- \frac{5}{4}$

Step 4.3

The graph is concave down on the interval $(- \infty, 0)$ because $f'' (- 2)$ is negative.

Concave down on $(- \infty, 0)$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(0, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

Replace the variable $x$ with $2$ in the expression.

$f'' (2) = \frac{10}{{(2)}^{3}}$

Step 5.2

Simplify the result.

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Step 5.2.1

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

$f'' (2) = \frac{10}{8}$

Step 5.2.2

Cancel the common factor of $10$ and $8$ .

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Step 5.2.2.1

Factor $2$ out of $10$ .

$f'' (2) = \frac{2 (5)}{8}$

Step 5.2.2.2

Cancel the common factors.

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Step 5.2.2.2.1

Factor $2$ out of $8$ .

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

Step 5.2.2.2.2

Cancel the common factor.

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

Step 5.2.2.2.3

Rewrite the expression.

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

Step 5.2.3

The final answer is $\frac{5}{4}$ .

$\frac{5}{4}$

Step 5.3

The graph is concave up on the interval $(0, \infty)$ because $f'' (2)$ is positive.

Concave up on $(0, \infty)$ since $f'' (x)$ is positive

Step 6

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

Concave down on $(- \infty, 0)$ since $f'' (x)$ is negative

Concave up on $(0, \infty)$ since $f'' (x)$ is positive

Step 7