Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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

Sagot :

Space

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]:

  1. f(x) = cxⁿ
  2. 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

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

Step 3: Find Tangent Slope

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

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

---

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

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

---

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

Unit: Differentiation