Calculus Examples

$x e^{- 3 x}$

Step 1

Write $x e^{- 3 x}$ as a function.

$f (x) = x e^{- 3 x}$

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

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 $f (x) = x$ and $g (x) = e^{- 3 x}$ .

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

Step 2.1.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 3 x$ .

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Step 2.1.1.2.1

To apply the Chain Rule, set $u$ as $- 3 x$ .

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

Step 2.1.1.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

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

Step 2.1.1.2.3

Replace all occurrences of $u$ with $- 3 x$ .

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

Step 2.1.1.3

Differentiate.

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Step 2.1.1.3.1

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

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

Step 2.1.1.3.2

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

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

Step 2.1.1.3.3

Simplify the expression.

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Step 2.1.1.3.3.1

Multiply $- 3$ by $1$ .

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

Step 2.1.1.3.3.2

Move $- 3$ to the left of $e^{- 3 x}$ .

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

Step 2.1.1.3.4

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

$x (- 3 e^{- 3 x}) + e^{- 3 x} \cdot 1$

Step 2.1.1.3.5

Multiply $e^{- 3 x}$ by $1$ .

$x (- 3 e^{- 3 x}) + e^{- 3 x}$

Step 2.1.1.4

Simplify.

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Step 2.1.1.4.1

Reorder terms.

$- 3 e^{- 3 x} x + e^{- 3 x}$

Step 2.1.1.4.2

Reorder factors in $- 3 e^{- 3 x} x + e^{- 3 x}$ .

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

Step 2.1.2

Find the second derivative.

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Step 2.1.2.1

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

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

Step 2.1.2.2

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

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Step 2.1.2.2.1

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

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

Step 2.1.2.2.2

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 $f (x) = x$ and $g (x) = e^{- 3 x}$ .

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

Step 2.1.2.2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 3 x$ .

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Step 2.1.2.2.3.1

To apply the Chain Rule, set $u_{1}$ as $- 3 x$ .

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

Step 2.1.2.2.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 2.1.2.2.3.3

Replace all occurrences of $u_{1}$ with $- 3 x$ .

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

Step 2.1.2.2.4

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

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

Step 2.1.2.2.5

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

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

Step 2.1.2.2.6

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

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

Step 2.1.2.2.7

Multiply $- 3$ by $1$ .

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

Step 2.1.2.2.8

Move $- 3$ to the left of $e^{- 3 x}$ .

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

Step 2.1.2.2.9

Multiply $e^{- 3 x}$ by $1$ .

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

Step 2.1.2.3

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

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Step 2.1.2.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 3 x$ .

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Step 2.1.2.3.1.1

To apply the Chain Rule, set $u_{2}$ as $- 3 x$ .

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

Step 2.1.2.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

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

Step 2.1.2.3.1.3

Replace all occurrences of $u_{2}$ with $- 3 x$ .

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

Step 2.1.2.3.2

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

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

Step 2.1.2.3.3

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

$- 3 (x (- 3 e^{- 3 x}) + e^{- 3 x}) + e^{- 3 x} (- 3 \cdot 1)$

Step 2.1.2.3.4

Multiply $- 3$ by $1$ .

$- 3 (x (- 3 e^{- 3 x}) + e^{- 3 x}) + e^{- 3 x} \cdot - 3$

Step 2.1.2.3.5

Move $- 3$ to the left of $e^{- 3 x}$ .

$- 3 (x (- 3 e^{- 3 x}) + e^{- 3 x}) - 3 e^{- 3 x}$

Step 2.1.2.4

Simplify.

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Step 2.1.2.4.1

Apply the distributive property.

$- 3 (x (- 3 e^{- 3 x})) - 3 e^{- 3 x} - 3 e^{- 3 x}$

Step 2.1.2.4.2

Combine terms.

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Step 2.1.2.4.2.1

Multiply $- 3$ by $- 3$ .

$9 (x (e^{- 3 x})) - 3 e^{- 3 x} - 3 e^{- 3 x}$

Step 2.1.2.4.2.2

Subtract $3 e^{- 3 x}$ from $- 3 e^{- 3 x}$ .

$9 x e^{- 3 x} - 6 e^{- 3 x}$

Step 2.1.2.4.3

Reorder terms.

$9 e^{- 3 x} x - 6 e^{- 3 x}$

Step 2.1.2.4.4

Reorder factors in $9 e^{- 3 x} x - 6 e^{- 3 x}$ .

$f'' (x) = 9 x e^{- 3 x} - 6 e^{- 3 x}$

Step 2.1.3

The second derivative of $f (x)$ with respect to $x$ is $9 x e^{- 3 x} - 6 e^{- 3 x}$ .

$9 x e^{- 3 x} - 6 e^{- 3 x}$

Step 2.2

Set the second derivative equal to $0$ then solve the equation $9 x e^{- 3 x} - 6 e^{- 3 x} = 0$ .

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Step 2.2.1

Set the second derivative equal to $0$ .

$9 x e^{- 3 x} - 6 e^{- 3 x} = 0$

Step 2.2.2

Factor $3 e^{- 3 x}$ out of $9 x e^{- 3 x} - 6 e^{- 3 x}$ .

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Step 2.2.2.1

Factor $3 e^{- 3 x}$ out of $9 x e^{- 3 x}$ .

$3 e^{- 3 x} (3 x) - 6 e^{- 3 x} = 0$

Step 2.2.2.2

Factor $3 e^{- 3 x}$ out of $- 6 e^{- 3 x}$ .

$3 e^{- 3 x} (3 x) + 3 e^{- 3 x} (- 2) = 0$

Step 2.2.2.3

Factor $3 e^{- 3 x}$ out of $3 e^{- 3 x} (3 x) + 3 e^{- 3 x} (- 2)$ .

$3 e^{- 3 x} (3 x - 2) = 0$

Step 2.2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$e^{- 3 x} = 0$

$3 x - 2 = 0$

Step 2.2.4

Set $e^{- 3 x}$ equal to $0$ and solve for $x$ .

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Step 2.2.4.1

Set $e^{- 3 x}$ equal to $0$ .

$e^{- 3 x} = 0$

Step 2.2.4.2

Solve $e^{- 3 x} = 0$ for $x$ .

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Step 2.2.4.2.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- 3 x}) = \ln (0)$

Step 2.2.4.2.2

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

Step 2.2.4.2.3

There is no solution for $e^{- 3 x} = 0$

No solution

Step 2.2.5

Set $3 x - 2$ equal to $0$ and solve for $x$ .

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Step 2.2.5.1

Set $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Step 2.2.5.2

Solve $3 x - 2 = 0$ for $x$ .

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Step 2.2.5.2.1

Add $2$ to both sides of the equation.

$3 x = 2$

Step 2.2.5.2.2

Divide each term in $3 x = 2$ by $3$ and simplify.

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Step 2.2.5.2.2.1

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Step 2.2.5.2.2.2

Simplify the left side.

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Step 2.2.5.2.2.2.1

Cancel the common factor of $3$ .

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Step 2.2.5.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Step 2.2.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{2}{3}$

Step 2.2.6

The final solution is all the values that make $3 e^{- 3 x} (3 x - 2) = 0$ true.

$x = \frac{2}{3}$

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.

$(- \infty, \frac{2}{3}) \cup (\frac{2}{3}, \infty)$

Step 5

Substitute any number from the interval $(- \infty, \frac{2}{3})$ 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.

$f'' (0) = 9 (0) \cdot e^{- 3 \cdot 0} - 6 e^{- 3 \cdot 0}$

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

Multiply $9$ by $0$ .

$f'' (0) = 0 \cdot e^{- 3 \cdot 0} - 6 e^{- 3 \cdot 0}$

Step 5.2.1.2

Multiply $- 3$ by $0$ .

$f'' (0) = 0 \cdot e^{0} - 6 e^{- 3 \cdot 0}$

Step 5.2.1.3

Anything raised to $0$ is $1$ .

$f'' (0) = 0 \cdot 1 - 6 e^{- 3 \cdot 0}$

Step 5.2.1.4

Multiply $0$ by $1$ .

$f'' (0) = 0 - 6 e^{- 3 \cdot 0}$

Step 5.2.1.5

Multiply $- 3$ by $0$ .

$f'' (0) = 0 - 6 e^{0}$

Step 5.2.1.6

Anything raised to $0$ is $1$ .

$f'' (0) = 0 - 6 \cdot 1$

Step 5.2.1.7

Multiply $- 6$ by $1$ .

$f'' (0) = 0 - 6$

Step 5.2.2

Subtract $6$ from $0$ .

$f'' (0) = - 6$

Step 5.2.3

The final answer is $- 6$ .

$- 6$

Step 5.3

The graph is concave down on the interval $(- \infty, \frac{2}{3})$ because $f'' (0)$ is negative.

Concave down on $(- \infty, \frac{2}{3})$ since $f'' (x)$ is negative

Step 6

Substitute any number from the interval $(\frac{2}{3}, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

Replace the variable $x$ with $3$ in the expression.

$f'' (3) = 9 (3) \cdot e^{- 3 \cdot 3} - 6 e^{- 3 \cdot 3}$

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

Multiply $9$ by $3$ .

$f'' (3) = 27 \cdot e^{- 3 \cdot 3} - 6 e^{- 3 \cdot 3}$

Step 6.2.1.2

Multiply $- 3$ by $3$ .

$f'' (3) = 27 \cdot e^{- 9} - 6 e^{- 3 \cdot 3}$

Step 6.2.1.3

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

$f'' (3) = 27 \cdot \frac{1}{e^{9}} - 6 e^{- 3 \cdot 3}$

Step 6.2.1.4

Combine $27$ and $\frac{1}{e^{9}}$ .

$f'' (3) = \frac{27}{e^{9}} - 6 e^{- 3 \cdot 3}$

Step 6.2.1.5

Multiply $- 3$ by $3$ .

$f'' (3) = \frac{27}{e^{9}} - 6 e^{- 9}$

Step 6.2.1.6

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

$f'' (3) = \frac{27}{e^{9}} - 6 \frac{1}{e^{9}}$

Step 6.2.1.7

Combine $- 6$ and $\frac{1}{e^{9}}$ .

$f'' (3) = \frac{27}{e^{9}} + \frac{- 6}{e^{9}}$

Step 6.2.1.8

Move the negative in front of the fraction.

$f'' (3) = \frac{27}{e^{9}} - \frac{6}{e^{9}}$

Step 6.2.2

Combine fractions.

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Step 6.2.2.1

Combine the numerators over the common denominator.

$f'' (3) = \frac{27 - 6}{e^{9}}$

Step 6.2.2.2

Subtract $6$ from $27$ .

$f'' (3) = \frac{21}{e^{9}}$

Step 6.2.3

The final answer is $\frac{21}{e^{9}}$ .

$\frac{21}{e^{9}}$

Step 6.3

The graph is concave up on the interval $(\frac{2}{3}, \infty)$ because $f'' (3)$ is positive.

Concave up on $(\frac{2}{3}, \infty)$ since $f'' (x)$ is positive

Step 7

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

Concave down on $(- \infty, \frac{2}{3})$ since $f'' (x)$ is negative

Concave up on $(\frac{2}{3}, \infty)$ since $f'' (x)$ is positive

Step 8