Calculus Examples

$y = 4 x - \ln (x)$

Step 1

Write $y = 4 x - \ln (x)$ as a function.

$f (x) = 4 x - \ln (x)$

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 $4 x - \ln (x)$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [- \ln (x)]$ .

$\frac{d}{d x} [4 x] + \frac{d}{d x} [- \ln (x)]$

Step 2.1.1.2

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

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Step 2.1.1.2.1

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

$4 \frac{d}{d x} [x] + \frac{d}{d x} [- \ln (x)]$

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 $n = 1$ .

$4 \cdot 1 + \frac{d}{d x} [- \ln (x)]$

Step 2.1.1.2.3

Multiply $4$ by $1$ .

$4 + \frac{d}{d x} [- \ln (x)]$

Step 2.1.1.3

Evaluate $\frac{d}{d x} [- \ln (x)]$ .

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Step 2.1.1.3.1

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

$4 - \frac{d}{d x} [\ln (x)]$

Step 2.1.1.3.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$4 - \frac{1}{x}$

Step 2.1.1.4

Reorder terms.

$f' (x) = - \frac{1}{x} + 4$

Step 2.1.2

Find the second derivative.

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Step 2.1.2.1

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

$\frac{d}{d x} [- \frac{1}{x}] + \frac{d}{d x} [4]$

Step 2.1.2.2

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

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Step 2.1.2.2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{1}{x}$ .

$- \frac{d}{d x} [\frac{1}{x}] + \frac{1}{x} \frac{d}{d x} [- 1] + \frac{d}{d x} [4]$

Step 2.1.2.2.2

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$- \frac{d}{d x} [x^{- 1}] + \frac{1}{x} \frac{d}{d x} [- 1] + \frac{d}{d x} [4]$

Step 2.1.2.2.3

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

$- - x^{- 2} + \frac{1}{x} \frac{d}{d x} [- 1] + \frac{d}{d x} [4]$

Step 2.1.2.2.4

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

$- - x^{- 2} + \frac{1}{x} \cdot 0 + \frac{d}{d x} [4]$

Step 2.1.2.2.5

Multiply $- 1$ by $- 1$ .

$1 x^{- 2} + \frac{1}{x} \cdot 0 + \frac{d}{d x} [4]$

Step 2.1.2.2.6

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

$x^{- 2} + \frac{1}{x} \cdot 0 + \frac{d}{d x} [4]$

Step 2.1.2.2.7

Multiply $\frac{1}{x}$ by $0$ .

$x^{- 2} + 0 + \frac{d}{d x} [4]$

Step 2.1.2.2.8

Add $x^{- 2}$ and $0$ .

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

Step 2.1.2.3

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$x^{- 2} + 0$

Step 2.1.2.4

Simplify.

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Step 2.1.2.4.1

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

$\frac{1}{x^{2}} + 0$

Step 2.1.2.4.2

Add $\frac{1}{x^{2}}$ and $0$ .

$f'' (x) = \frac{1}{x^{2}}$

Step 2.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{1}{x^{2}}$ .

$\frac{1}{x^{2}}$

Step 2.2

Set the second derivative equal to $0$ then solve the equation $\frac{1}{x^{2}} = 0$ .

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Step 2.2.1

Set the second derivative equal to $0$ .

$\frac{1}{x^{2}} = 0$

Step 2.2.2

Set the numerator equal to zero.

$1 = 0$

Step 2.2.3

Since $1 \neq 0$ , there are no solutions.

No solution

Step 3

Find the domain of $f (x) = 4 x - \ln (x)$ .

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Step 3.1

Set the argument in $\ln (x)$ greater than $0$ to find where the expression is defined.

$x > 0$

Step 3.2

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

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Step 4

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

$(0, \infty)$

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.

$f'' (2) = \frac{1}{{(2)}^{2}}$

Step 5.2

Simplify the result.

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Step 5.2.1

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

$f'' (2) = \frac{1}{4}$

Step 5.2.2

The final answer is $\frac{1}{4}$ .

$\frac{1}{4}$

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