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Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

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BuyFindarrow_forward

Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

Find the points where tangents to the graph of the equation in Problem 21 are horizontal.

To determine

To calculate: The point at which the curve defined by x2+4x3y2+6=0 has horizontal tangents.

Explanation

Given Information:

The provided equation of the curve is:

x2+4x3y2+6=0

Formula used:

The slope of tangent to a curve y=f(x) at point (x,y) is given by the derivative of the curve at that point.

Calculation:

Consider the provided equation of curve:

x2+4x3y2+6=0

Now, find the derivative dydx from x2+4x3y2+6=0 by taking the derivative term by term on both sides of the equation as:

ddx(x2)+ddx(4x)ddx(3y2)+ddx(6)=02x+46ydydx+0=0

Solve for dydx as:

2x+46ydy

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Chapter 11 Solutions

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