36. Find an equation of the tangent line to the curve at the given point.y = Vx – x , (1,0)56. For what value of x does the graph of f(x) = ex – 2x have a horizontal tangent?

Given equation y=x14-xDifferenciate with respective xdydx=14x14-1…

Answered: 36. Find an equation of the tangent…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

36. Find an equation of the tangent line to the curve at the given point.
y = Vx – x , (1,0)
56. For what value of x does the graph of f(x) = ex – 2x have a horizontal tangent?

Expert Solution

Step 1

$G i v e n e q u a t i o n y = x^{\frac{1}{4}} - x D i f f e r e n c i a t e w i t h r e s p e c t i v e x \frac{d y}{d x} = \frac{1}{4} x^{\frac{1}{4} - 1} - 1 \frac{d y}{d x} = \frac{1}{4} x^{- 3 / 4} - 1 {\frac{d y}{d x}}_{(x, y) = (1, 0)} = \frac{1}{4} {(1)}^{- 3 / 4} - 1 {\frac{d y}{d x}}_{(x, y) = (1, 0)} = \frac{1}{4} - 1 {\frac{d y}{d x}}_{(x, y) = (1, 0)} = - \frac{3}{4} t h e r e f o r e e q u a t i o n o f \tan g e n t i s o f t h e f o r m y - y_{1} = m (x - x_{}) . w h e r e m i s s l o p e o f l i n e . t h a t i s \frac{d y}{d x} . t h e r e f o r e y - 0 = \frac{- 3}{4} (x - 1) y = \frac{- 3}{4} (x - 1) 4 y = - 3 (x - 1) 4 y = - 3 x + 3 3 x + 4 y = 3$