Finding the equation of the normal line at

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Homework Help Overview

The problem involves finding the equation of the normal line to the graph of the function y=3x4 - 1/x at the point where x = 1. The discussion centers around the calculation of the slope of the tangent line and the subsequent determination of the normal line's equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the derivative to find the slope of the tangent line, with one participant noting a potential error in the derivative of 1/x. There is also a focus on the negative reciprocal of the tangent slope to find the normal line's slope.

Discussion Status

The discussion is ongoing, with participants verifying calculations and clarifying derivative concepts. Some guidance has been offered regarding the derivative, and there is acknowledgment of a minor mistake that does not affect the final outcome.

Contextual Notes

Participants express uncertainty about the correctness of their calculations and seek verification of their results, indicating a focus on understanding rather than simply obtaining the answer.

meeklobraca
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Homework Statement



Find the equation, in standard form, of the normal line drawn to the graph of y=3x4 - 1/x at x = 1



Homework Equations





The Attempt at a Solution



-I figured out the slope of the tangent to be 12x3 + 1, then at x = 1 the slope is 13
-SO the slope of the normal being the negative reciprocal of 13 led me to get -1/13
-Plug that into the y - y1 = m(x - x1) equation and I get x +13y - 27 = 0 for the equation of the normal.

Im a little gun shy since the numbers seem abnormally large, so I would love if someone could verify this, or if I am wrong, point me in the right direction of being right!

Cheers.
 
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meeklobraca said:

Homework Statement



Find the equation, in standard form, of the normal line drawn to the graph of y=3x4 - 1/x at x = 1



Homework Equations





The Attempt at a Solution



-I figured out the slope of the tangent to be 12x3 + 1, then at x = 1 the slope is 13
-SO the slope of the normal being the negative reciprocal of 13 led me to get -1/13
-Plug that into the y - y1 = m(x - x1) equation and I get x +13y - 27 = 0 for the equation of the normal.

Im a little gun shy since the numbers seem abnormally large, so I would love if someone could verify this, or if I am wrong, point me in the right direction of being right!

Cheers.

You are right that the derivative of [tex]3 {x^4}[/tex] is [tex]12{x^3}[/tex]

But what is the derivative ot 1/x again?
 
You made a minor mistake in your derivative, it should be 12x³ + 1/x², not 12x³ + 1. However, this doesn't affect the slope at x=1, and the rest of your derivation (including the answer) is correct.
 
YUP i see where I made the mistake. Instead of subtracting the 1 from the x-1 I added. But your right it doesn't affect the outcome. But thanks guys!
 

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