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Sagot :
Answer:
The equation of the tangent line is given by the following equation:
[tex]\displaystyle y - \frac{1}{e} = \frac{-1}{e} \bigg( x - 1 \bigg)[/tex]
General Formulas and Concepts:
Algebra I
Point-Slope Form: y - y₁ = m(x - x₁)
- x₁ - x coordinate
- y₁ - y coordinate
- m - slope
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]:
[tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Rule [Basic Power Rule]:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Chain Rule]:
[tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Step-by-step explanation:
*Note:
Recall that the definition of the derivative is the slope of the tangent line.
Step 1: Define
Identify given.
[tex]\displaystylef(x) = e^{-x} \\x = -1[/tex]
Step 2: Differentiate
- [Function] Apply Exponential Differentiation [Derivative Rule - Chain Rule]:
[tex]\displaystyle f'(x) = e^{-x}(-x)'[/tex] - [Derivative] Rewrite [Derivative Rule - Multiplied Constant]:
[tex]\displaystyle f'(x) = -e^{-x}(x)'[/tex] - [Derivative] Apply Derivative Rule [Derivative Rule - Basic Power Rule]:
[tex]\displaystyle f'(x) = -e^{-x}[/tex]
Step 3: Find Tangent Slope
- [Derivative] Substitute in x = 1:
[tex]\displaystyle f'(1) = -e^{-1}[/tex] - Rewrite:
[tex]\displaystyle f'(1) = \frac{-1}{e}[/tex]
∴ the slope of the tangent line is equal to [tex]\displaystyle \frac{-1}{e}[/tex].
Step 4: Find Equation
- [Function] Substitute in x = 1:
[tex]\displaystyle f(1) = e^{-1}[/tex] - Rewrite:
[tex]\displaystyle f(1) = \frac{1}{e}[/tex]
∴ our point is equal to [tex]\displaystyle \bigg( 1, \frac{1}{e} \bigg)[/tex].
Substituting in our variables we found into the point-slope form general equation, we get our final answer of:
[tex]\displaystyle \boxed{ y - \frac{1}{e} = \frac{-1}{e} \bigg( x - 1 \bigg) }[/tex]
∴ we have our final answer.
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Learn more about derivatives: https://brainly.com/question/27163229
Learn more about calculus: https://brainly.com/question/23558817
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Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
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