Question

F(x) = e^-x . Find the equation of the tangent to f(x) at x=-1​

Answer 1

Answer:

The equation of the tangent line is given by the following equation:

$\displaystyle y - \frac{1}{e} = \frac{-1}{e} \bigg( x - 1 \bigg)$

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]:
$\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Rule [Basic Power Rule]:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

*Note:

Recall that the definition of the derivative is the slope of the tangent line.

Step 1: Define

Identify given.

$\displaystylef(x) = e^{-x} \\x = -1$

Step 2: Differentiate

1. [Function] Apply Exponential Differentiation [Derivative Rule - Chain Rule]:
   $\displaystyle f'(x) = e^{-x}(-x)'$
2. [Derivative] Rewrite [Derivative Rule - Multiplied Constant]:
   $\displaystyle f'(x) = -e^{-x}(x)'$
3. [Derivative] Apply Derivative Rule [Derivative Rule - Basic Power Rule]:
   $\displaystyle f'(x) = -e^{-x}$

Step 3: Find Tangent Slope

1. [Derivative] Substitute in x = 1:
   $\displaystyle f'(1) = -e^{-1}$
2. Rewrite:
   $\displaystyle f'(1) = \frac{-1}{e}$

∴ the slope of the tangent line is equal to  $\displaystyle \frac{-1}{e}$.

Step 4: Find Equation

1. [Function] Substitute in x = 1:
   $\displaystyle f(1) = e^{-1}$
2. Rewrite:
   $\displaystyle f(1) = \frac{1}{e}$

∴ our point is equal to  $\displaystyle \bigg( 1, \frac{1}{e} \bigg)$.

Substituting in our variables we found into the point-slope form general equation, we get our final answer of:

$\displaystyle \boxed{ y - \frac{1}{e} = \frac{-1}{e} \bigg( x - 1 \bigg) }$

∴ we have our final answer.

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Learn more about derivatives: brainly.com/question/27163229

Learn more about calculus: brainly.com/question/23558817

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation