Question

I have a function f(x)=sqrt(2x+1). I need to find point c such that f'(c)=1.....

Answer 1

So I suppose I have to differentiate (? find the general formula for the derivative?), but I'm not sure how to apply the basic differentiation rules here

Answer 2

ok so if \[\sqrt{2x+1}=(2x+1)^\frac{ 1 }{ 2 }\] then using differentiation I get \[\frac{ 1 }{ 2 }(2x+1)^\frac{ -1 }{ 2 }\]

Answer 3

which is \[\frac{ 1 }{ 2 }*\frac{ 1 }{ (2x+1)^\frac{ 1 }{ 2 } }\]

Answer 4

You forgot your chain rule.

http://tutorial.math.lamar.edu/Classes/CalcI/ChainRule.aspx

Your derivative f'(x) = 1/(2x+1)

Answer 5

You want a point c such that f'(c) = 1

So if 1 = 1/(2c+1)

2c + 1 = 1

2c = 0

c = 0

f'(0) = 1

Answer 6

verification http://www.wolframalpha.com/input/?i=derivative+of+%282x%2B1%29%5E.5

Answer 7

veification f'(0) = 1 http://www.wolframalpha.com/input/?i=1%2Fsqrt%281%2B2+x%29+%3D+1

Answer 8

\[f(x) = (2x+1)^{1/2}\]\[u=2x+1\]\[du/dx=2\]\[y=u^{1/2}\]\[dy/dx=dy/du*du/dx\]\[y'=\frac{1}{2\sqrt u}2\]\[y'=1/\sqrt{2x+1}\]

Answer 9

The 1/2 and the 2 cancel out

Answer 10

wait, but how do you get rid of the square root?

Answer 11

I did not rationalize the denominator, the problem was to solve for a point c that yields f'(c) = 1, not to simplify the denominator of my derivative.

Answer 12

If you wanted to:\[f'(x)=\frac{1}{\sqrt{2x+1}}*\frac{\sqrt{2x+1}}{\sqrt{2x+1}}\]\[f'(x)=\frac{\sqrt{2x+1}}{2x+1}\]

Answer 13

oh so it's just a question of 0 being an obvious solution?

Answer 14

Well we want a point x such that f'(x) = 1\[f'(x)=1/\sqrt{2x+1}\]\[f'(x)=1\]Substitute\[1=1/\sqrt{2x+1}\]Cross multiply\[\sqrt{2x+1}=1\]Square both sides\[2x+1=1^2\]\[2x+1=1\]\[2x=0\]\[x=0/2\]\[x=0\]

Answer 15

So f'(0) = 1

Answer 16

ok, thanks stamp