Answered: (a) What is the slope of the line…

College Algebra

7th Edition

ISBN:9781305115545

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter1: Equations And Graphs

Section1.3: Lines

Problem 3E

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Question

(a) What is the slope of the line tangent to y = xe^-x at x = 1?

(b) Write the equation of the line tangent to the graph of y = xe^-x at x = 1.

Expert Solution

Step 1

We have to find:

$(a)$ slope of line tangent to the curve $y = x e^{- x}$ at $x = 1$

$(b)$ equation of line tangent to the graph of $y = x e^{- x}$ at $x = 1$ .

$(a)$ We know that slope $(m)$ of line and $\frac{d y}{d x}$ is same

so differentiating the curve,

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d (x e^{- x})}{d x} \\ = & x \frac{d e^{- x}}{d x} + e^{- x} \frac{d x}{d x} \\ = & x e^{- x} (\frac{d (- x)}{d x}) + e^{- x} \times 1 \\ = & x e^{- x} (- 1) + e^{- x} \\ = & - x e^{- x} + e^{- x} \\ = & e^{- x} (- x + 1) \end{array}$

We have used following differentiation formula,

$\begin{array}{rcl} \frac{d e^{x}}{d x} & = & e^{x} \\ \frac{d e^{f (x)}}{d x} & = & e^{f (x)} \frac{d f (x)}{d x} \\ \frac{d x^{n}}{d x} & = & n x^{n - 1} \end{array}$

So finding slope at $x = 1$ ,

$\begin{array}{rcl} m & = & {\frac{d y}{d x}}_{a t x = 1} \\ = & e^{- x} (- x + 1) \\ = & e^{- 1} (- 1 + 1) \\ = & e^{- 1} \times 0 \\ = & 0 \end{array}$

Hence, slope of line tangent is $0$ .

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