Calculus Examples

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

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

[0, 2]

$[0, 2]$

Step 1

f

$f$ is continuous on the interval

[a, b]

$[a, b]$ and differentiable on

(a, b)

$(a, b)$ , then at least one real number

c

$c$ exists in the interval

(a, b)

$(a, b)$ such that

$f' (c) = \frac{f (b) - f a}{b - a}$ . The mean value theorem expresses the relationship between the slope of the tangent to the curve at

$x = c$ and the slope of the line through the points

$(a, f (a))$ and

$(b, f (b))$ .

$f (x)$ is continuous on

$[a, b]$

and if

$f (x)$ differentiable on

$(a, b)$ ,

then there exists at least one point,

$c$ in

$[a, b]$ :

$f' (c) = \frac{f (b) - f a}{b - a}$ .

Step 2

Check if

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

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Step 2.1

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.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 2.2

$f (x)$ is continuous on

$[0, 2]$ .

The function is continuous.

Step 3

Find the derivative.

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Step 3.1

Find the first derivative.

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Step 3.1.1

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 3.1.1.1

To apply the Chain Rule, set

$u$ as

$- 2 x$ .

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

Step 3.1.1.2

Differentiate using the Exponential Rule which states that

$\frac{d}{d u} [a^{u}]$ is

$a^{u} \ln (a)$ where

$a$ =

$e$ .

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

Step 3.1.1.3

Replace all occurrences of

$u$ with

$- 2 x$ .

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

Step 3.1.2

Differentiate.

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Step 3.1.2.1

Since

$- 2$ is constant with respect to

$x$ , the derivative of

$- 2 x$ with respect to

$x$ is

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

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

Step 3.1.2.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 3.1.2.3

Simplify the expression.

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Step 3.1.2.3.1

Multiply

$- 2$ by

$1$ .

$e^{- 2 x} \cdot - 2$

Step 3.1.2.3.2

Move

$- 2$ to the left of

$e^{- 2 x}$ .

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

Step 3.2

The first derivative of

$f (x)$ with respect to

$x$ is

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

$- 2 e^{- 2 x}$

Step 4

Find if the derivative is continuous on

$(0, 2)$ .

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Step 4.1

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.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4.2

$f' (x)$ is continuous on

$(0, 2)$ .

The function is continuous.

Step 5

The function is differentiable on

$(0, 2)$ because the derivative is continuous on

$(0, 2)$ .

The function is differentiable.

Step 6

$f (x)$ satisfies the two conditions for the mean value theorem. It is continuous on

$[0, 2]$ and differentiable on

$(0, 2)$ .

$f (x)$ is continuous on

$[0, 2]$ and differentiable on

$(0, 2)$ .

Step 7

Evaluate

$f (a)$ from the interval

$[0, 2]$ .

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Step 7.1

Replace the variable

$x$ with

$0$ in the expression.

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

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply

$- 2$ by

$0$ .

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

Step 7.2.2

Anything raised to

$0$ is

$1$ .

$f (0) = 1$

Step 7.2.3

The final answer is

$1$ .

$1$

Step 8

Evaluate

$f (b)$ from the interval

$[0, 2]$ .

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Step 8.1

Replace the variable

$x$ with

$2$ in the expression.

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

Step 8.2

Simplify the result.

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Step 8.2.1

Multiply

$- 2$ by

$2$ .

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

Step 8.2.2

Rewrite the expression using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

$f (2) = \frac{1}{e^{4}}$

Step 8.2.3

The final answer is

$\frac{1}{e^{4}}$ .

$\frac{1}{e^{4}}$

Step 9

Solve

$- 2 e^{- 2 x} = \frac{- (f (b) + f (a))}{- (b + a)}$ for

$x$ .

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

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Step 9.1

Simplify

$\frac{(\frac{1}{e^{4}}) - (1)}{(2) - (0)}$ .

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Step 9.1.1

Multiply the numerator and denominator of the fraction by

$e^{4}$ .

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Step 9.1.1.1

Multiply

$\frac{\frac{1}{e^{4}} - (1)}{2 - (0)}$ by

$\frac{e^{4}}{e^{4}}$ .

$- 2 e^{- 2 x} = \frac{e^{4}}{e^{4}} \cdot \frac{\frac{1}{e^{4}} - (1)}{2 - (0)}$

Step 9.1.1.2

Combine.

$- 2 e^{- 2 x} = \frac{e^{4} (\frac{1}{e^{4}} - (1))}{e^{4} (2 - (0))}$

Step 9.1.2

Apply the distributive property.

$- 2 e^{- 2 x} = \frac{e^{4} \frac{1}{e^{4}} + e^{4} (- (1))}{e^{4} \cdot 2 + e^{4} (- (0))}$

Step 9.1.3

Cancel the common factor of

$e^{4}$ .

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Step 9.1.3.1

Cancel the common factor.

$- 2 e^{- 2 x} = \frac{e^{4} \frac{1}{e^{4}} + e^{4} (- (1))}{e^{4} \cdot 2 + e^{4} (- (0))}$

Step 9.1.3.2

Rewrite the expression.

$- 2 e^{- 2 x} = \frac{1 + e^{4} (- (1))}{e^{4} \cdot 2 + e^{4} (- (0))}$

Step 9.1.4

Simplify the numerator.

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Step 9.1.4.1

Rewrite

$1$ as

$1^{2}$ .

$- 2 e^{- 2 x} = \frac{1^{2} + e^{4} (- (1))}{e^{4} \cdot 2 + e^{4} (- (0))}$

Step 9.1.4.2

Rewrite

$e^{4} \cdot (1)$ as

${(e^{2} \cdot 1)}^{2}$ .

$- 2 e^{- 2 x} = \frac{1^{2} - {(e^{2} \cdot 1)}^{2}}{e^{4} \cdot 2 + e^{4} (- (0))}$

Step 9.1.4.3

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = 1$ and

$b = e^{2} \cdot 1$ .

$- 2 e^{- 2 x} = \frac{(1 + e^{2} \cdot 1) (1 - (e^{2} \cdot 1))}{e^{4} \cdot 2 + e^{4} (- (0))}$

Step 9.1.4.4

Simplify.

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Step 9.1.4.4.1

Multiply

$e^{2}$ by

$1$ .

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) (1 - (e^{2} \cdot 1))}{e^{4} \cdot 2 + e^{4} (- (0))}$

Step 9.1.4.4.2

Rewrite

$1$ as

$1^{2}$ .

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) (1^{2} - e^{2} \cdot 1)}{e^{4} \cdot 2 + e^{4} (- (0))}$

Step 9.1.4.4.3

Rewrite

$e^{2} \cdot 1$ as

${(e \cdot 1)}^{2}$ .

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) (1^{2} - {(e \cdot 1)}^{2})}{e^{4} \cdot 2 + e^{4} (- (0))}$

Step 9.1.4.4.4

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = 1$ and

$b = e \cdot 1$ .

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) ((1 + e \cdot 1) (1 - (e \cdot 1)))}{e^{4} \cdot 2 + e^{4} (- (0))}$

Step 9.1.4.4.5

Simplify.

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Step 9.1.4.4.5.1

Multiply

$e$ by

$1$ .

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) ((1 + e) (1 - (e \cdot 1)))}{e^{4} \cdot 2 + e^{4} (- (0))}$

Step 9.1.4.4.5.2

Multiply

$e$ by

$1$ .

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) ((1 + e) (1 - e))}{e^{4} \cdot 2 + e^{4} (- (0))}$

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) (1 + e) (1 - e)}{e^{4} \cdot 2 + e^{4} (- (0))}$

Step 9.1.5

Simplify the denominator.

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Step 9.1.5.1

Factor

$e^{4}$ out of

$e^{4} \cdot 2 + e^{4} (- (0))$ .

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) (1 + e) (1 - e)}{e^{4} (2 - (0))}$

Step 9.1.5.2

Multiply

$- 1$ by

$0$ .

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) (1 + e) (1 - e)}{e^{4} (2 + 0)}$

Step 9.1.5.3

Add

$2$ and

$0$ .

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) (1 + e) (1 - e)}{e^{4} \cdot 2}$

Step 9.1.6

Move

$2$ to the left of

$e^{4}$ .

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) (1 + e) (1 - e)}{2 e^{4}}$

Step 9.2

Divide each term in

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) (1 + e) (1 - e)}{2 e^{4}}$ by

$- 2$ and simplify.

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Step 9.2.1

Divide each term in

$- 2 e^{- 2 x} = \frac{(1 + e^{2}) (1 + e) (1 - e)}{2 e^{4}}$ by

$- 2$ .

$\frac{- 2 e^{- 2 x}}{- 2} = \frac{\frac{(1 + e^{2}) (1 + e) (1 - e)}{2 e^{4}}}{- 2}$

Step 9.2.2

Simplify the left side.

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Step 9.2.2.1

Cancel the common factor of

$- 2$ .

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Step 9.2.2.1.1

Cancel the common factor.

$\frac{- 2 e^{- 2 x}}{- 2} = \frac{\frac{(1 + e^{2}) (1 + e) (1 - e)}{2 e^{4}}}{- 2}$

Step 9.2.2.1.2

Divide

$e^{- 2 x}$ by

$1$ .

$e^{- 2 x} = \frac{\frac{(1 + e^{2}) (1 + e) (1 - e)}{2 e^{4}}}{- 2}$

Step 9.2.3

Simplify the right side.

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Step 9.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$e^{- 2 x} = \frac{(1 + e^{2}) (1 + e) (1 - e)}{2 e^{4}} \cdot \frac{1}{- 2}$

Step 9.2.3.2

Combine.

$e^{- 2 x} = \frac{(1 + e^{2}) (1 + e) (1 - e) \cdot 1}{2 e^{4} \cdot - 2}$

Step 9.2.3.3

Simplify the expression.

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Step 9.2.3.3.1

Multiply

$1 + e^{2}$ by

$1$ .

$e^{- 2 x} = \frac{(1 + e^{2}) (1 + e) (1 - e)}{2 e^{4} \cdot - 2}$

Step 9.2.3.3.2

Multiply

$- 2$ by

$2$ .

$e^{- 2 x} = \frac{(1 + e^{2}) (1 + e) (1 - e)}{- 4 e^{4}}$

Step 9.2.3.3.3

Move the negative in front of the fraction.

$e^{- 2 x} = - \frac{(1 + e^{2}) (1 + e) (1 - e)}{4 e^{4}}$

Step 9.3

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

$\ln (e^{- 2 x}) = \ln (- \frac{(1 + e^{2}) (1 + e) (1 - e)}{4 e^{4}})$

Step 9.4

The equation cannot be solved because

$\ln (- \frac{(1 + e^{2}) (1 + e) (1 - e)}{4 e^{4}})$ is undefined.

Undefined

Step 9.5

There is no solution for

$e^{- 2 x} = - \frac{(1 + e^{2}) (1 + e) (1 - e)}{4 e^{4}}$

No solution

Step 10

There are no solution, so there is no

$x$ value where the tangent line is parallel to the line that passes through the end points

$a = 0$ and

$b = 2$ .

No x value found where the tangent line at x is parallel to the line that passes through the end points

$a = 0$ and

$b = 2$

Step 11