Calculus Examples

$f (x) = 35 x - 2 e^{x}$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $35 x - 2 e^{x}$ with respect to $x$ is $\frac{d}{d x} [35 x] + \frac{d}{d x} [- 2 e^{x}]$ .

$\frac{d}{d x} [35 x] + \frac{d}{d x} [- 2 e^{x}]$

Step 1.2

Evaluate $\frac{d}{d x} [35 x]$ .

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Step 1.2.1

Since $35$ is constant with respect to $x$ , the derivative of $35 x$ with respect to $x$ is $35 \frac{d}{d x} [x]$ .

$35 \frac{d}{d x} [x] + \frac{d}{d x} [- 2 e^{x}]$

Step 1.2.2

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

$35 \cdot 1 + \frac{d}{d x} [- 2 e^{x}]$

Step 1.2.3

Multiply $35$ by $1$ .

$35 + \frac{d}{d x} [- 2 e^{x}]$

Step 1.3

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

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Step 1.3.1

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

$35 - 2 \frac{d}{d x} [e^{x}]$

Step 1.3.2

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

$35 - 2 e^{x}$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $35 - 2 e^{x}$ with respect to $x$ is $\frac{d}{d x} [35] + \frac{d}{d x} [- 2 e^{x}]$ .

$f'' (x) = \frac{d}{d x} (35) + \frac{d}{d x} (- 2 e^{x})$

Step 2.1.2

Since $35$ is constant with respect to $x$ , the derivative of $35$ with respect to $x$ is $0$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

Subtract $2 e^{x}$ from $0$ .

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

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$35 - 2 e^{x} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

By the Sum Rule, the derivative of $35 x - 2 e^{x}$ with respect to $x$ is $\frac{d}{d x} [35 x] + \frac{d}{d x} [- 2 e^{x}]$ .

$\frac{d}{d x} [35 x] + \frac{d}{d x} [- 2 e^{x}]$

Step 4.1.2

Evaluate $\frac{d}{d x} [35 x]$ .

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Step 4.1.2.1

Since $35$ is constant with respect to $x$ , the derivative of $35 x$ with respect to $x$ is $35 \frac{d}{d x} [x]$ .

$35 \frac{d}{d x} [x] + \frac{d}{d x} [- 2 e^{x}]$

Step 4.1.2.2

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

$35 \cdot 1 + \frac{d}{d x} [- 2 e^{x}]$

Step 4.1.2.3

Multiply $35$ by $1$ .

$35 + \frac{d}{d x} [- 2 e^{x}]$

Step 4.1.3

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

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Step 4.1.3.1

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

$35 - 2 \frac{d}{d x} [e^{x}]$

Step 4.1.3.2

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

$f' (x) = 35 - 2 e^{x}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $35 - 2 e^{x}$ .

$35 - 2 e^{x}$

Step 5

Set the first derivative equal to $0$ then solve the equation $35 - 2 e^{x} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$35 - 2 e^{x} = 0$

Step 5.2

Subtract $35$ from both sides of the equation.

$- 2 e^{x} = - 35$

Step 5.3

Divide each term in $- 2 e^{x} = - 35$ by $- 2$ and simplify.

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Step 5.3.1

Divide each term in $- 2 e^{x} = - 35$ by $- 2$ .

$\frac{- 2 e^{x}}{- 2} = \frac{- 35}{- 2}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $- 2$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{- 2 e^{x}}{- 2} = \frac{- 35}{- 2}$

Step 5.3.2.1.2

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

$e^{x} = \frac{- 35}{- 2}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Dividing two negative values results in a positive value.

$e^{x} = \frac{35}{2}$

Step 5.4

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

$\ln (e^{x}) = \ln (\frac{35}{2})$

Step 5.5

Expand the left side.

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Step 5.5.1

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

$x \ln (e) = \ln (\frac{35}{2})$

Step 5.5.2

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

$x \cdot 1 = \ln (\frac{35}{2})$

Step 5.5.3

Multiply $x$ by $1$ .

$x = \ln (\frac{35}{2})$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

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.

Step 7

Critical points to evaluate.

$x = \ln (\frac{35}{2})$

Step 8

Evaluate the second derivative at $x = \ln (\frac{35}{2})$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 2 e^{\ln (\frac{35}{2})}$

Step 9

Evaluate the second derivative.

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Step 9.1

Exponentiation and log are inverse functions.

$- 2 (\frac{35}{2})$

Step 9.2

Cancel the common factor of $2$ .

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Step 9.2.1

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{35}{2}$

Step 9.2.2

Cancel the common factor.

$2 \cdot - 1 \frac{35}{2}$

Step 9.2.3

Rewrite the expression.

$- 1 \cdot 35$

Step 9.3

Multiply $- 1$ by $35$ .

$- 35$

Step 10

$x = \ln (\frac{35}{2})$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \ln (\frac{35}{2})$ is a local maximum

Step 11

Find the y-value when $x = \ln (\frac{35}{2})$ .

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Step 11.1

Replace the variable $x$ with $\ln (\frac{35}{2})$ in the expression.

$f (\ln (\frac{35}{2})) = 35 (\ln (\frac{35}{2})) - 2 e^{\ln (\frac{35}{2})}$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Simplify $35 (\ln (\frac{35}{2}))$ by moving $35$ inside the logarithm.

$f (\ln (\frac{35}{2})) = \ln ({(\frac{35}{2})}^{35}) - 2 e^{\ln (\frac{35}{2})}$

Step 11.2.1.2

Apply the product rule to $\frac{35}{2}$ .

$f (\ln (\frac{35}{2})) = \ln (\frac{35^{35}}{2^{35}}) - 2 e^{\ln (\frac{35}{2})}$

Step 11.2.1.3

Raise $2$ to the power of $35$ .

$f (\ln (\frac{35}{2})) = \ln (\frac{35^{35}}{34359738368}) - 2 e^{\ln (\frac{35}{2})}$

Step 11.2.1.4

Exponentiation and log are inverse functions.

$f (\ln (\frac{35}{2})) = \ln (\frac{35^{35}}{34359738368}) - 2 (\frac{35}{2})$

Step 11.2.1.5

Cancel the common factor of $2$ .

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Step 11.2.1.5.1

Factor $2$ out of $- 2$ .

$f (\ln (\frac{35}{2})) = \ln (\frac{35^{35}}{34359738368}) + 2 (- 1) (\frac{35}{2})$

Step 11.2.1.5.2

Cancel the common factor.

$f (\ln (\frac{35}{2})) = \ln (\frac{35^{35}}{34359738368}) + 2 \cdot (- 1 (\frac{35}{2}))$

Step 11.2.1.5.3

Rewrite the expression.

$f (\ln (\frac{35}{2})) = \ln (\frac{35^{35}}{34359738368}) - 1 \cdot 35$

Step 11.2.1.6

Multiply $- 1$ by $35$ .

$f (\ln (\frac{35}{2})) = \ln (\frac{35^{35}}{34359738368}) - 35$

Step 11.2.2

The final answer is $\ln (\frac{35^{35}}{34359738368}) - 35$ .

$y = \ln (\frac{35^{35}}{34359738368}) - 35$

Step 12

These are the local extrema for $f (x) = 35 x - 2 e^{x}$ .

$(\ln (\frac{35}{2}), \ln (\frac{35^{35}}{34359738368}) - 35)$ is a local maxima

Step 13