Calculus Examples

$e^{x} (x - 2)$

Step 1

Write $e^{x} (x - 2)$ as a function.

$f (x) = e^{x} (x - 2)$

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

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 $f (x) = e^{x}$ and $g (x) = x - 2$ .

$f' (x) = e^{x} \frac{d}{d x} (x - 2) + (x - 2) \frac{d}{d x} (e^{x})$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

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

$f' (x) = e^{x} (\frac{d}{d x} (x) + \frac{d}{d x} (- 2)) + (x - 2) \frac{d}{d x} (e^{x})$

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 $n = 1$ .

$f' (x) = e^{x} (1 + \frac{d}{d x} (- 2)) + (x - 2) \frac{d}{d x} (e^{x})$

Step 2.1.2.3

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$f' (x) = e^{x} (1 + 0) + (x - 2) \frac{d}{d x} (e^{x})$

Step 2.1.2.4

Simplify the expression.

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Step 2.1.2.4.1

Add $1$ and $0$ .

$f' (x) = e^{x} \cdot 1 + (x - 2) \frac{d}{d x} (e^{x})$

Step 2.1.2.4.2

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

$f' (x) = e^{x} + (x - 2) \frac{d}{d x} (e^{x})$

Step 2.1.3

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

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

Step 2.1.4

Simplify.

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Step 2.1.4.1

Apply the distributive property.

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

Step 2.1.4.2

Subtract $2 e^{x}$ from $e^{x}$ .

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

Step 2.1.4.3

Reorder terms.

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

Step 2.1.4.4

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

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

Step 2.2

Find the second derivative.

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Step 2.2.1

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

$f'' (x) = \frac{d}{d x} (x e^{x}) + \frac{d}{d x} (- e^{x})$

Step 2.2.2

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

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Step 2.2.2.1

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

$f'' (x) = x \frac{d}{d x} (e^{x}) + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (- e^{x})$

Step 2.2.2.2

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

$f'' (x) = x e^{x} + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (- e^{x})$

Step 2.2.2.3

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

$f'' (x) = x e^{x} + e^{x} \cdot 1 + \frac{d}{d x} (- e^{x})$

Step 2.2.2.4

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

$f'' (x) = x e^{x} + e^{x} + \frac{d}{d x} (- e^{x})$

Step 2.2.3

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

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Step 2.2.3.1

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

$f'' (x) = x e^{x} + e^{x} - \frac{d}{d x} e^{x}$

Step 2.2.3.2

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

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

Step 2.2.4

Simplify.

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Step 2.2.4.1

Combine terms.

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Step 2.2.4.1.1

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

$f'' (x) = x e^{x} + 0$

Step 2.2.4.1.2

Add $x e^{x}$ and $0$ .

$f'' (x) = x e^{x}$

Step 2.2.4.2

Reorder the factors of $x e^{x}$ .

$f'' (x) = e^{x} x$

Step 2.2.4.3

Reorder factors in $e^{x} x$ .

$f'' (x) = x e^{x}$

Step 2.3

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

$x e^{x}$

Step 3

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

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Step 3.1

Set the second derivative equal to $0$ .

$x e^{x} = 0$

Step 3.2

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

$x = 0$

$e^{x} = 0$

Step 3.3

Set $x$ equal to $0$ .

$x = 0$

Step 3.4

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

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Step 3.4.1

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

$e^{x} = 0$

Step 3.4.2

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

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Step 3.4.2.1

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

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

Step 3.4.2.2

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

Undefined

Step 3.4.2.3

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

No solution

Step 3.5

The final solution is all the values that make $x e^{x} = 0$ true.

$x = 0$

Step 4

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

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Step 4.1

Substitute $0$ in $f (x) = e^{x} (x - 2)$ to find the value of $y$ .

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Step 4.1.1

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

$f (0) = e^{0} ((0) - 2)$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Anything raised to $0$ is $1$ .

$f (0) = 1 ((0) - 2)$

Step 4.1.2.2

Multiply $(0) - 2$ by $1$ .

$f (0) = (0) - 2$

Step 4.1.2.3

Subtract $2$ from $0$ .

$f (0) = - 2$

Step 4.1.2.4

The final answer is $- 2$ .

$- 2$

Step 4.2

The point found by substituting $0$ in $f (x) = e^{x} (x - 2)$ is $(0, - 2)$ . This point can be an inflection point.

$(0, - 2)$

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.

$f'' (- 0.1) = (- 0.1) \cdot e^{- 0.1}$

Step 6.2

Simplify the result.

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Step 6.2.1

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

$f'' (- 0.1) = - 0.1 \cdot \frac{1}{e^{0.1}}$

Step 6.2.2

Combine $- 0.1$ and $\frac{1}{e^{0.1}}$ .

$f'' (- 0.1) = \frac{- 0.1}{e^{0.1}}$

Step 6.2.3

Move the negative in front of the fraction.

$f'' (- 0.1) = - \frac{0.1}{e^{0.1}}$

Step 6.2.4

Replace $e$ with an approximation.

$f'' (- 0.1) = - \frac{0.1}{{2.71828182}^{0.1}}$

Step 6.2.5

Raise $2.71828182$ to the power of $0.1$ .

$f'' (- 0.1) = - \frac{0.1}{1.10517091}$

Step 6.2.6

Divide $0.1$ by $1.10517091$ .

$f'' (- 0.1) = - 1 \cdot 0.09048374$

Step 6.2.7

Multiply $- 1$ by $0.09048374$ .

$f'' (- 0.1) = - 0.09048374$

Step 6.2.8

The final answer is $- 0.09048374$ .

$- 0.09048374$

Step 6.3

At $- 0.1$ , the second derivative is $- 0.09048374$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 0)$

Decreasing 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.

$f'' (0.1) = (0.1) \cdot e^{0.1}$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $0.1$ by $e^{0.1}$ .

$f'' (0.1) = 0.11051709$

Step 7.2.2

The final answer is $0.11051709$ .

$0.11051709$

Step 7.3

At $0.1$ , the second derivative is $0.11051709$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

Increasing 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 $(0, - 2)$ .

$(0, - 2)$

Step 9