Calculus Examples

$f (x) = 10 x - 10 e^{- x}$

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 $10 x - 10 e^{- x}$ with respect to $x$ is $\frac{d}{d x} [10 x] + \frac{d}{d x} [- 10 e^{- x}]$ .

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

Step 1.1.1.2

Evaluate $\frac{d}{d x} [10 x]$ .

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Step 1.1.1.2.1

Since $10$ is constant with respect to $x$ , the derivative of $10 x$ with respect to $x$ is $10 \frac{d}{d x} [x]$ .

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

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 $n = 1$ .

$10 \cdot 1 + \frac{d}{d x} [- 10 e^{- x}]$

Step 1.1.1.2.3

Multiply $10$ by $1$ .

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

Step 1.1.1.3

Evaluate $\frac{d}{d x} [- 10 e^{- x}]$ .

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Step 1.1.1.3.1

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10 e^{- x}$ with respect to $x$ is $- 10 \frac{d}{d x} [e^{- x}]$ .

$10 - 10 \frac{d}{d x} [e^{- x}]$

Step 1.1.1.3.2

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 1.1.1.3.2.1

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

$10 - 10 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x])$

Step 1.1.1.3.2.2

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

$10 - 10 (e^{u} \frac{d}{d x} [- x])$

Step 1.1.1.3.2.3

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

$10 - 10 (e^{- x} \frac{d}{d x} [- x])$

Step 1.1.1.3.3

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

$10 - 10 (e^{- x} (- \frac{d}{d x} [x]))$

Step 1.1.1.3.4

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

$10 - 10 (e^{- x} (- 1 \cdot 1))$

Step 1.1.1.3.5

Multiply $- 1$ by $1$ .

$10 - 10 (e^{- x} \cdot - 1)$

Step 1.1.1.3.6

Move $- 1$ to the left of $e^{- x}$ .

$10 - 10 (- 1 \cdot e^{- x})$

Step 1.1.1.3.7

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$10 - 10 (- e^{- x})$

Step 1.1.1.3.8

Multiply $- 1$ by $- 10$ .

$10 + 10 e^{- x}$

Step 1.1.1.4

Reorder terms.

$f' (x) = 10 e^{- x} + 10$

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

By the Sum Rule, the derivative of $10 e^{- x} + 10$ with respect to $x$ is $\frac{d}{d x} [10 e^{- x}] + \frac{d}{d x} [10]$ .

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

Step 1.1.2.2

Evaluate $\frac{d}{d x} [10 e^{- x}]$ .

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Step 1.1.2.2.1

Since $10$ is constant with respect to $x$ , the derivative of $10 e^{- x}$ with respect to $x$ is $10 \frac{d}{d x} [e^{- x}]$ .

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

Step 1.1.2.2.2

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 1.1.2.2.2.1

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

$10 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x]) + \frac{d}{d x} [10]$

Step 1.1.2.2.2.2

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

$10 (e^{u} \frac{d}{d x} [- x]) + \frac{d}{d x} [10]$

Step 1.1.2.2.2.3

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

$10 (e^{- x} \frac{d}{d x} [- x]) + \frac{d}{d x} [10]$

Step 1.1.2.2.3

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

$10 (e^{- x} (- \frac{d}{d x} [x])) + \frac{d}{d x} [10]$

Step 1.1.2.2.4

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

$10 (e^{- x} (- 1 \cdot 1)) + \frac{d}{d x} [10]$

Step 1.1.2.2.5

Multiply $- 1$ by $1$ .

$10 (e^{- x} \cdot - 1) + \frac{d}{d x} [10]$

Step 1.1.2.2.6

Move $- 1$ to the left of $e^{- x}$ .

$10 (- 1 \cdot e^{- x}) + \frac{d}{d x} [10]$

Step 1.1.2.2.7

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$10 (- e^{- x}) + \frac{d}{d x} [10]$

Step 1.1.2.2.8

Multiply $- 1$ by $10$ .

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

Step 1.1.2.3

Differentiate using the Constant Rule.

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Step 1.1.2.3.1

Since $10$ is constant with respect to $x$ , the derivative of $10$ with respect to $x$ is $0$ .

$- 10 e^{- x} + 0$

Step 1.1.2.3.2

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

$f'' (x) = - 10 e^{- x}$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $- 10 e^{- x}$ .

$- 10 e^{- x}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $- 10 e^{- x} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$- 10 e^{- x} = 0$

Step 1.2.2

Divide each term in $- 10 e^{- x} = 0$ by $- 10$ and simplify.

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Step 1.2.2.1

Divide each term in $- 10 e^{- x} = 0$ by $- 10$ .

$\frac{- 10 e^{- x}}{- 10} = \frac{0}{- 10}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $- 10$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{- 10 e^{- x}}{- 10} = \frac{0}{- 10}$

Step 1.2.2.2.1.2

Divide $e^{- x}$ by $1$ .

$e^{- x} = \frac{0}{- 10}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Divide $0$ by $- 10$ .

$e^{- x} = 0$

Step 1.2.3

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

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

Step 1.2.4

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

Undefined

Step 1.2.5

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

No solution

Step 2

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 3

The graph is concave down because the second derivative is negative.

The graph is concave down

Step 4