Find the x coordinate of the stationary point of the following curves


by studentxlol
Tags: coordinate, curves, point, stationary
studentxlol
studentxlol is offline
#1
Nov28-11, 04:24 AM
P: 39
1. The problem statement, all variables and given/known data

Find dy/dx and determine the exact x coordinate of the stationary point for:

(a) y=(4x^2+1)^5

(b) y=x^2/lnx

2. Relevant equations


3. The attempt at a solution

(a) y=(4x^2+1)^5

dy/dx=40x(4x^2+1)^4

40x(4x^2+1)^4=0

Find x... How?

(b) y=x^2/lnx

dy/dx=2xlnx-x^2 1/x / (lnx)^2

2xlnx-x^2 1/x / (lnx)^2=0

Find x... How?
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grzz
grzz is offline
#2
Nov28-11, 04:38 AM
P: 939
Quote Quote by studentxlol View Post
40x(4x^2+1)^4=0

Find x... How?
re 1st prob:

Then either 40x = 0 or (4x^2+1)^4 = 0.

and solve the above two equations.
HallsofIvy
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#3
Nov28-11, 06:58 AM
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Thanks
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P: 38,898
You are aware that [itex]x^2/x= x[/itex] aren't you?

[itex]y= x^2/ln(x)[/itex]: [itex]y'= (2xln(x)- x)/(ln(x))^2= 0[/itex]
Use parentheses! What you wrote was [itex]y'= 2x ln(x)- (x/(ln(x))^2)= 0[/itex].

Multiply both sides of the equation by [itex](ln(x))^2[/itex]
and you are left with 2x ln(x)- x= x(2ln(x)- 1)= 0. Can you solve that?


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