Calculus Examples

$20 e^{x} - e^{2 x}$

Step 1

Write $20 e^{x} - e^{2 x}$ as a function.

$f (x) = 20 e^{x} - e^{2 x}$

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

By the Sum Rule, the derivative of $20 e^{x} - e^{2 x}$ with respect to $x$ is $\frac{d}{d x} [20 e^{x}] + \frac{d}{d x} [- e^{2 x}]$ .

$\frac{d}{d x} [20 e^{x}] + \frac{d}{d x} [- e^{2 x}]$

Step 2.1.2

Evaluate $\frac{d}{d x} [20 e^{x}]$ .

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Step 2.1.2.1

Since $20$ is constant with respect to $x$ , the derivative of $20 e^{x}$ with respect to $x$ is $20 \frac{d}{d x} [e^{x}]$ .

$20 \frac{d}{d x} [e^{x}] + \frac{d}{d x} [- e^{2 x}]$

Step 2.1.2.2

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

$20 e^{x} + \frac{d}{d x} [- e^{2 x}]$

Step 2.1.3

Evaluate $\frac{d}{d x} [- e^{2 x}]$ .

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Step 2.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- e^{2 x}$ with respect to $x$ is $- \frac{d}{d x} [e^{2 x}]$ .

$20 e^{x} - \frac{d}{d x} [e^{2 x}]$

Step 2.1.3.2

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 $g (x) = 2 x$ .

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Step 2.1.3.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$20 e^{x} - (\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x])$

Step 2.1.3.2.2

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

$20 e^{x} - (e^{u} \frac{d}{d x} [2 x])$

Step 2.1.3.2.3

Replace all occurrences of $u$ with $2 x$ .

$20 e^{x} - (e^{2 x} \frac{d}{d x} [2 x])$

Step 2.1.3.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$20 e^{x} - (e^{2 x} (2 \frac{d}{d x} [x]))$

Step 2.1.3.4

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

$20 e^{x} - (e^{2 x} (2 \cdot 1))$

Step 2.1.3.5

Multiply $2$ by $1$ .

$20 e^{x} - (e^{2 x} \cdot 2)$

Step 2.1.3.6

Move $2$ to the left of $e^{2 x}$ .

$20 e^{x} - (2 \cdot e^{2 x})$

Step 2.1.3.7

Multiply $2$ by $- 1$ .

$f' (x) = 20 e^{x} - 2 e^{2 x}$

Step 2.2

Find the second derivative.

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Step 2.2.1

By the Sum Rule, the derivative of $20 e^{x} - 2 e^{2 x}$ with respect to $x$ is $\frac{d}{d x} [20 e^{x}] + \frac{d}{d x} [- 2 e^{2 x}]$ .

$\frac{d}{d x} [20 e^{x}] + \frac{d}{d x} [- 2 e^{2 x}]$

Step 2.2.2

Evaluate $\frac{d}{d x} [20 e^{x}]$ .

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Step 2.2.2.1

Since $20$ is constant with respect to $x$ , the derivative of $20 e^{x}$ with respect to $x$ is $20 \frac{d}{d x} [e^{x}]$ .

$20 \frac{d}{d x} [e^{x}] + \frac{d}{d x} [- 2 e^{2 x}]$

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 $a$ = $e$ .

$20 e^{x} + \frac{d}{d x} [- 2 e^{2 x}]$

Step 2.2.3

Evaluate $\frac{d}{d x} [- 2 e^{2 x}]$ .

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Step 2.2.3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 e^{2 x}$ with respect to $x$ is $- 2 \frac{d}{d x} [e^{2 x}]$ .

$20 e^{x} - 2 \frac{d}{d x} [e^{2 x}]$

Step 2.2.3.2

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 $g (x) = 2 x$ .

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Step 2.2.3.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$20 e^{x} - 2 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x])$

Step 2.2.3.2.2

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

$20 e^{x} - 2 (e^{u} \frac{d}{d x} [2 x])$

Step 2.2.3.2.3

Replace all occurrences of $u$ with $2 x$ .

$20 e^{x} - 2 (e^{2 x} \frac{d}{d x} [2 x])$

Step 2.2.3.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$20 e^{x} - 2 (e^{2 x} (2 \frac{d}{d x} [x]))$

Step 2.2.3.4

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

$20 e^{x} - 2 (e^{2 x} (2 \cdot 1))$

Step 2.2.3.5

Multiply $2$ by $1$ .

$20 e^{x} - 2 (e^{2 x} \cdot 2)$

Step 2.2.3.6

Move $2$ to the left of $e^{2 x}$ .

$20 e^{x} - 2 (2 \cdot e^{2 x})$

Step 2.2.3.7

Multiply $2$ by $- 2$ .

$f'' (x) = 20 e^{x} - 4 e^{2 x}$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $20 e^{x} - 4 e^{2 x}$ .

$20 e^{x} - 4 e^{2 x}$

Step 3

Set the second derivative equal to $0$ then solve the equation $20 e^{x} - 4 e^{2 x} = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$20 e^{x} - 4 e^{2 x} = 0$

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

Rewrite $e^{2 x}$ as ${(e^{x})}^{2}$ .

$20 e^{x} - 4 {(e^{x})}^{2} = 0$

Step 3.2.2

Let $u = e^{x}$ . Substitute $u$ for all occurrences of $e^{x}$ .

$20 u - 4 u^{2} = 0$

Step 3.2.3

Factor $4 u$ out of $20 u - 4 u^{2}$ .

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Step 3.2.3.1

Factor $4 u$ out of $20 u$ .

$4 u (5) - 4 u^{2} = 0$

Step 3.2.3.2

Factor $4 u$ out of $- 4 u^{2}$ .

$4 u (5) + 4 u (- u) = 0$

Step 3.2.3.3

Factor $4 u$ out of $4 u (5) + 4 u (- u)$ .

$4 u (5 - u) = 0$

Step 3.2.4

Replace all occurrences of $u$ with $e^{x}$ .

$4 e^{x} (5 - e^{x}) = 0$

Step 3.3

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

$e^{x} = 0$

$5 - e^{x} = 0$

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

Set $5 - e^{x}$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $5 - e^{x}$ equal to $0$ .

$5 - e^{x} = 0$

Step 3.5.2

Solve $5 - e^{x} = 0$ for $x$ .

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Step 3.5.2.1

Subtract $5$ from both sides of the equation.

$- e^{x} = - 5$

Step 3.5.2.2

Divide each term in $- e^{x} = - 5$ by $- 1$ and simplify.

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Step 3.5.2.2.1

Divide each term in $- e^{x} = - 5$ by $- 1$ .

$\frac{- e^{x}}{- 1} = \frac{- 5}{- 1}$

Step 3.5.2.2.2

Simplify the left side.

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Step 3.5.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{e^{x}}{1} = \frac{- 5}{- 1}$

Step 3.5.2.2.2.2

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

$e^{x} = \frac{- 5}{- 1}$

Step 3.5.2.2.3

Simplify the right side.

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Step 3.5.2.2.3.1

Divide $- 5$ by $- 1$ .

$e^{x} = 5$

Step 3.5.2.3

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

$\ln (e^{x}) = \ln (5)$

Step 3.5.2.4

Expand the left side.

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Step 3.5.2.4.1

Expand $\ln (e^{x})$ by moving $x$ outside the logarithm.

$x \ln (e) = \ln (5)$

Step 3.5.2.4.2

The natural logarithm of $e$ is $1$ .

$x \cdot 1 = \ln (5)$

Step 3.5.2.4.3

Multiply $x$ by $1$ .

$x = \ln (5)$

Step 3.6

The final solution is all the values that make $4 e^{x} (5 - e^{x}) = 0$ true.

$x = \ln (5)$

Step 4

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

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Step 4.1

Substitute $\ln (5)$ in $f (x) = 20 e^{x} - e^{2 x}$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $\ln (5)$ in the expression.

$f (\ln (5)) = 20 e^{\ln (5)} - e^{2 (\ln (5))}$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

Exponentiation and log are inverse functions.

$f (\ln (5)) = 20 \cdot 5 - e^{2 (\ln (5))}$

Step 4.1.2.1.2

Multiply $20$ by $5$ .

$f (\ln (5)) = 100 - e^{2 (\ln (5))}$

Step 4.1.2.1.3

Simplify $2 (\ln (5))$ by moving $2$ inside the logarithm.

$f (\ln (5)) = 100 - e^{\ln (5^{2})}$

Step 4.1.2.1.4

Exponentiation and log are inverse functions.

$f (\ln (5)) = 100 - 5^{2}$

Step 4.1.2.1.5

Raise $5$ to the power of $2$ .

$f (\ln (5)) = 100 - 1 \cdot 25$

Step 4.1.2.1.6

Multiply $- 1$ by $25$ .

$f (\ln (5)) = 100 - 25$

Step 4.1.2.2

Subtract $25$ from $100$ .

$f (\ln (5)) = 75$

Step 4.1.2.3

The final answer is $75$ .

$75$

Step 4.2

The point found by substituting $\ln (5)$ in $f (x) = 20 e^{x} - e^{2 x}$ is $(\ln (5), 75)$ . This point can be an inflection point.

$(\ln (5), 75)$

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, \ln (5)) \cup (\ln (5), \infty)$

Step 6

Substitute a value from the interval $(- \infty, \ln (5))$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $1.50943791$ in the expression.

$f'' (1.50943791) = 20 e^{1.50943791} - 4 e^{2 (1.50943791)}$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $2$ by $1.50943791$ .

$f'' (1.50943791) = 20 e^{1.50943791} - 4 e^{3.01887582}$

Step 6.2.2

The final answer is $20 e^{1.50943791} - 4 e^{3.01887582}$ .

$20 e^{1.50943791} - 4 e^{3.01887582}$

Step 6.3

At $1.50943791$ , the second derivative is $8.61066649$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, \ln (5))$ .

Increasing on $(- \infty, \ln (5))$ since $f'' (x) > 0$

Step 7

Substitute a value from the interval $(\ln (5), \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $1.70943791$ in the expression.

$f'' (1.70943791) = 20 e^{1.70943791} - 4 e^{2 (1.70943791)}$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $2$ by $1.70943791$ .

$f'' (1.70943791) = 20 e^{1.70943791} - 4 e^{3.41887582}$

Step 7.2.2

The final answer is $20 e^{1.70943791} - 4 e^{3.41887582}$ .

$20 e^{1.70943791} - 4 e^{3.41887582}$

Step 7.3

At $1.70943791$ , the second derivative is $- 11.623184$ . Since this is negative, the second derivative is decreasing on the interval $(\ln (5), \infty)$

Decreasing on $(\ln (5), \infty)$ 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 $(\ln (5), 75)$ .

$(\ln (5), 75)$

Step 9