Calculus Examples

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

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 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 $f (x) = x^{2} - 3$ and $g (x) = e^{- x}$ .

$f' (x) = (x^{2} - 3) \frac{d}{d x} (e^{- x}) + e^{- x} \frac{d}{d x} (x^{2} - 3)$

Step 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 $g (x) = - x$ .

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Step 1.1.2.1

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

$f' (x) = (x^{2} - 3) (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x^{2} - 3)$

Step 1.1.2.2

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

$f' (x) = (x^{2} - 3) (e^{u} \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x^{2} - 3)$

Step 1.1.2.3

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

$f' (x) = (x^{2} - 3) (e^{- x} \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x^{2} - 3)$

Step 1.1.3

Differentiate.

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Step 1.1.3.1

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

$f' (x) = (x^{2} - 3) e^{- x} (- \frac{d}{d x} x) + e^{- x} \frac{d}{d x} (x^{2} - 3)$

Step 1.1.3.2

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

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

Step 1.1.3.3

Simplify the expression.

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Step 1.1.3.3.1

Multiply $- 1$ by $1$ .

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

Step 1.1.3.3.2

Move $- 1$ to the left of $(x^{2} - 3) e^{- x}$ .

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

Step 1.1.3.3.3

Rewrite $- 1 (x^{2} - 3)$ as $- (x^{2} - 3)$ .

$f' (x) = - (x^{2} - 3) e^{- x} + e^{- x} \frac{d}{d x} (x^{2} - 3)$

Step 1.1.3.4

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

$f' (x) = - (x^{2} - 3) e^{- x} + e^{- x} (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 3))$

Step 1.1.3.5

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

$f' (x) = - (x^{2} - 3) e^{- x} + e^{- x} (2 x + \frac{d}{d x} (- 3))$

Step 1.1.3.6

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

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

Step 1.1.3.7

Add $2 x$ and $0$ .

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

Step 1.1.4

Simplify.

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Step 1.1.4.1

Apply the distributive property.

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

Step 1.1.4.2

Apply the distributive property.

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

Step 1.1.4.3

Multiply $- 1$ by $- 3$ .

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

Step 1.1.4.4

Reorder terms.

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

Step 1.1.4.5

Reorder factors in $- e^{- x} x^{2} + 2 e^{- x} x + 3 e^{- x}$ .

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

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- x^{2} e^{- x} + 2 x e^{- x} + 3 e^{- x}$ .

$- x^{2} e^{- x} + 2 x e^{- x} + 3 e^{- x}$

Step 2

Set the first derivative equal to $0$ then solve the equation $- x^{2} e^{- x} + 2 x e^{- x} + 3 e^{- x} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

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

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

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

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Step 2.2.1.1

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

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

Step 2.2.1.2

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

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

Step 2.2.1.3

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

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

Step 2.2.1.4

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

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

Step 2.2.1.5

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

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

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor by grouping.

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Step 2.2.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 3 = - 3$ and whose sum is $b = 2$ .

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Step 2.2.2.1.1.1

Factor $2$ out of $2 x$ .

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

Step 2.2.2.1.1.2

Rewrite $2$ as $- 1$ plus $3$

$e^{- x} (- x^{2} + (- 1 + 3) x + 3) = 0$

Step 2.2.2.1.1.3

Apply the distributive property.

$e^{- x} (- x^{2} - 1 x + 3 x + 3) = 0$

Step 2.2.2.1.2

Factor out the greatest common factor from each group.

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Step 2.2.2.1.2.1

Group the first two terms and the last two terms.

$e^{- x} ((- x^{2} - 1 x) + 3 x + 3) = 0$

Step 2.2.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$e^{- x} (x (- x - 1) - 3 (- x - 1)) = 0$

Step 2.2.2.1.3

Factor the polynomial by factoring out the greatest common factor, $- x - 1$ .

$e^{- x} ((- x - 1) (x - 3)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$e^{- x} (- x - 1) (x - 3) = 0$

Step 2.3

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

$e^{- x} = 0$

$- x - 1 = 0$

$x - 3 = 0$

Step 2.4

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

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Step 2.4.1

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

$e^{- x} = 0$

Step 2.4.2

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

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Step 2.4.2.1

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

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

Step 2.4.2.2

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

Undefined

Step 2.4.2.3

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

No solution

Step 2.5

Set $- x - 1$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $- x - 1$ equal to $0$ .

$- x - 1 = 0$

Step 2.5.2

Solve $- x - 1 = 0$ for $x$ .

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Step 2.5.2.1

Add $1$ to both sides of the equation.

$- x = 1$

Step 2.5.2.2

Divide each term in $- x = 1$ by $- 1$ and simplify.

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Step 2.5.2.2.1

Divide each term in $- x = 1$ by $- 1$ .

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

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 2.5.2.2.2.2

Divide $x$ by $1$ .

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

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Divide $1$ by $- 1$ .

$x = - 1$

Step 2.6

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

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Step 2.6.1

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

$x - 3 = 0$

Step 2.6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.7

The final solution is all the values that make $e^{- x} (- x - 1) (x - 3) = 0$ true.

$x = - 1, 3$

Step 3

The values which make the derivative equal to $0$ are $- 1, 3$ .

$- 1, 3$

Step 4

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the derivative $0$ or undefined.

$(- \infty, - 1) \cup (- 1, 3) \cup (3, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 1)$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

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

$f' (- 2) = - {(- 2)}^{2} e^{- (- 2)} + 2 (- 2) \cdot e^{- (- 2)} + 3 e^{- (- 2)}$

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

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

$f' (- 2) = - 1 \cdot (4 e^{- (- 2)}) + 2 (- 2) \cdot e^{- (- 2)} + 3 e^{- (- 2)}$

Step 5.2.1.2

Multiply $- 1$ by $4$ .

$f' (- 2) = - 4 e^{- (- 2)} + 2 (- 2) \cdot e^{- (- 2)} + 3 e^{- (- 2)}$

Step 5.2.1.3

Multiply $- 1$ by $- 2$ .

$f' (- 2) = - 4 e^{2} + 2 (- 2) \cdot e^{- (- 2)} + 3 e^{- (- 2)}$

Step 5.2.1.4

Multiply $2$ by $- 2$ .

$f' (- 2) = - 4 e^{2} - 4 \cdot e^{- (- 2)} + 3 e^{- (- 2)}$

Step 5.2.1.5

Multiply $- 1$ by $- 2$ .

$f' (- 2) = - 4 e^{2} - 4 \cdot e^{2} + 3 e^{- (- 2)}$

Step 5.2.1.6

Multiply $- 1$ by $- 2$ .

$f' (- 2) = - 4 e^{2} - 4 e^{2} + 3 e^{2}$

Step 5.2.2

Simplify by adding terms.

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Step 5.2.2.1

Subtract $4 e^{2}$ from $- 4 e^{2}$ .

$f' (- 2) = - 8 e^{2} + 3 e^{2}$

Step 5.2.2.2

Add $- 8 e^{2}$ and $3 e^{2}$ .

$f' (- 2) = - 5 e^{2}$

Step 5.2.3

The final answer is $- 5 e^{2}$ .

$- 5 e^{2}$

Step 5.3

At $x = - 2$ the derivative is $- 5 e^{2}$ . Since this is negative, the function is decreasing on $(- \infty, - 1)$ .

Decreasing on $(- \infty, - 1)$ since $f' (x) < 0$

Step 6

Substitute a value from the interval $(- 1, 3)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $1$ in the expression.

$f' (1) = - {(1)}^{2} e^{- (1)} + 2 (1) \cdot e^{- (1)} + 3 e^{- (1)}$

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

One to any power is one.

$f' (1) = - 1 \cdot (1 e^{- (1)}) + 2 (1) \cdot e^{- (1)} + 3 e^{- (1)}$

Step 6.2.1.2

Multiply $- 1$ by $1$ .

$f' (1) = - 1 e^{- (1)} + 2 (1) \cdot e^{- (1)} + 3 e^{- (1)}$

Step 6.2.1.3

Multiply $- 1$ by $1$ .

$f' (1) = - 1 e^{- 1} + 2 (1) \cdot e^{- (1)} + 3 e^{- (1)}$

Step 6.2.1.4

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

$f' (1) = - 1 \frac{1}{e} + 2 (1) \cdot e^{- (1)} + 3 e^{- (1)}$

Step 6.2.1.5

Rewrite $- 1 \frac{1}{e}$ as $- \frac{1}{e}$ .

$f' (1) = - \frac{1}{e} + 2 (1) \cdot e^{- (1)} + 3 e^{- (1)}$

Step 6.2.1.6

Multiply $2$ by $1$ .

$f' (1) = - \frac{1}{e} + 2 \cdot e^{- (1)} + 3 e^{- (1)}$

Step 6.2.1.7

Multiply $- 1$ by $1$ .

$f' (1) = - \frac{1}{e} + 2 \cdot e^{- 1} + 3 e^{- (1)}$

Step 6.2.1.8

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

$f' (1) = - \frac{1}{e} + 2 \cdot \frac{1}{e} + 3 e^{- (1)}$

Step 6.2.1.9

Combine $2$ and $\frac{1}{e}$ .

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

Step 6.2.1.10

Multiply $- 1$ by $1$ .

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

Step 6.2.1.11

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

$f' (1) = - \frac{1}{e} + \frac{2}{e} + 3 (\frac{1}{e})$

Step 6.2.1.12

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

$f' (1) = - \frac{1}{e} + \frac{2}{e} + \frac{3}{e}$

Step 6.2.2

Combine fractions.

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Step 6.2.2.1

Combine the numerators over the common denominator.

$f' (1) = \frac{- 1 + 2 + 3}{e}$

Step 6.2.2.2

Simplify by adding numbers.

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Step 6.2.2.2.1

Add $- 1$ and $2$ .

$f' (1) = \frac{1 + 3}{e}$

Step 6.2.2.2.2

Add $1$ and $3$ .

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

Step 6.2.3

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

$\frac{4}{e}$

Step 6.3

At $x = 1$ the derivative is $\frac{4}{e}$ . Since this is positive, the function is increasing on $(- 1, 3)$ .

Increasing on $(- 1, 3)$ since $f' (x) > 0$

Step 7

Substitute a value from the interval $(3, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

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

$f' (4) = - {(4)}^{2} e^{- (4)} + 2 (4) \cdot e^{- (4)} + 3 e^{- (4)}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Raise $4$ to the power of $2$ .

$f' (4) = - 1 \cdot (16 e^{- (4)}) + 2 (4) \cdot e^{- (4)} + 3 e^{- (4)}$

Step 7.2.1.2

Multiply $- 1$ by $16$ .

$f' (4) = - 16 e^{- (4)} + 2 (4) \cdot e^{- (4)} + 3 e^{- (4)}$

Step 7.2.1.3

Multiply $- 1$ by $4$ .

$f' (4) = - 16 e^{- 4} + 2 (4) \cdot e^{- (4)} + 3 e^{- (4)}$

Step 7.2.1.4

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

$f' (4) = - 16 \frac{1}{e^{4}} + 2 (4) \cdot e^{- (4)} + 3 e^{- (4)}$

Step 7.2.1.5

Combine $- 16$ and $\frac{1}{e^{4}}$ .

$f' (4) = \frac{- 16}{e^{4}} + 2 (4) \cdot e^{- (4)} + 3 e^{- (4)}$

Step 7.2.1.6

Move the negative in front of the fraction.

$f' (4) = - \frac{16}{e^{4}} + 2 (4) \cdot e^{- (4)} + 3 e^{- (4)}$

Step 7.2.1.7

Multiply $2$ by $4$ .

$f' (4) = - \frac{16}{e^{4}} + 8 \cdot e^{- (4)} + 3 e^{- (4)}$

Step 7.2.1.8

Multiply $- 1$ by $4$ .

$f' (4) = - \frac{16}{e^{4}} + 8 \cdot e^{- 4} + 3 e^{- (4)}$

Step 7.2.1.9

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

$f' (4) = - \frac{16}{e^{4}} + 8 \cdot \frac{1}{e^{4}} + 3 e^{- (4)}$

Step 7.2.1.10

Combine $8$ and $\frac{1}{e^{4}}$ .

$f' (4) = - \frac{16}{e^{4}} + \frac{8}{e^{4}} + 3 e^{- (4)}$

Step 7.2.1.11

Multiply $- 1$ by $4$ .

$f' (4) = - \frac{16}{e^{4}} + \frac{8}{e^{4}} + 3 e^{- 4}$

Step 7.2.1.12

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

$f' (4) = - \frac{16}{e^{4}} + \frac{8}{e^{4}} + 3 (\frac{1}{e^{4}})$

Step 7.2.1.13

Combine $3$ and $\frac{1}{e^{4}}$ .

$f' (4) = - \frac{16}{e^{4}} + \frac{8}{e^{4}} + \frac{3}{e^{4}}$

Step 7.2.2

Combine fractions.

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Step 7.2.2.1

Combine the numerators over the common denominator.

$f' (4) = \frac{- 16 + 8 + 3}{e^{4}}$

Step 7.2.2.2

Simplify the expression.

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Step 7.2.2.2.1

Add $- 16$ and $8$ .

$f' (4) = \frac{- 8 + 3}{e^{4}}$

Step 7.2.2.2.2

Add $- 8$ and $3$ .

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

Step 7.2.2.2.3

Move the negative in front of the fraction.

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

Step 7.2.3

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

$- \frac{5}{e^{4}}$

Step 7.3

At $x = 4$ the derivative is $- \frac{5}{e^{4}}$ . Since this is negative, the function is decreasing on $(3, \infty)$ .

Decreasing on $(3, \infty)$ since $f' (x) < 0$

Step 8

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- 1, 3)$

Decreasing on: $(- \infty, - 1), (3, \infty)$

Step 9