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The First Principles

  • Thread starter roam
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[SOLVED] The First Principles

Hello!
This is my question:

1) Write down the definition of ƒ'(3), where [tex]f(x) = \sqrt{x + 1}[/tex]




2) Use the definition to evaluate ƒ'(3).



This isn't an exercise from a book etc.. so I don't know what's the correct solution to this question but here's my attempt:

By the first principles( [tex]f'(x) = lim h\rightarrow0 \frac{f(x+h) - f(x)}{h}[/tex] ):
∴ [tex]\frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}[/tex]

So would the definition of ƒ'(3) be like this:

[tex]f'(x) = \frac{\sqrt{3+h} - \sqrt{3+1}}{h}[/tex]

Simplifying gives:
[tex]= 1[/tex]


But another time I didn't use the first principles:

[tex]f(x) = \sqrt{x + 1}[/tex]

[tex]1/2(x+1)^-1/2[/tex]

[tex]\frac{1}{1/2.\sqrt{x+1}}[/tex]

[tex]\frac{1}{1/2.\sqrt{3+1}} = 1[/tex]

(?) - "?"

I got the same answer using both methods (=1). Am I on the right track? What else do I need to show?


Thank you. :smile:

 
Last edited:

Answers and Replies

737
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Please recheck what you get for this:

[tex]f'(x) = \frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}[/tex]

Remember you need to substitute x+h for x.
 
1,265
11
You mean like:
[tex]\frac{(x+h) - \sqrt{x+1}}{h}[/tex]

[tex]\frac{(3+h) - \sqrt{3+1}}{h}[/tex]

[tex]3 - 2[/tex] , → [tex] = 1[/tex]

But is the rest of my above working correct? Did I show everything that was needed?
What would be the right solution to my question? :rolleyes:
 
Gib Z
Homework Helper
3,344
4
No thats not what he meant :( Recheck your working, if f(x) = sqrt( x+1), what is f(x+h) ? Its not sqrt(x+h).
 
HallsofIvy
Science Advisor
Homework Helper
41,733
893
Hello!
This is my question:

1) Write down the definition of ƒ'(3), where [tex]f(x) = \sqrt{x + 1}[/tex]




2) Use the definition to evaluate ƒ'(3).



This isn't an exercise from a book etc.. so I don't know what's the correct solution to this question but here's my attempt:

By the first principles( [tex]f'(x) = lim h\rightarrow0 \frac{f(x+h) - f(x)}{h}[/tex] ):
∴ [tex]\frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}[/tex]

So would the definition of ƒ'(3) be like this:

[tex]f'(x) = \frac{\sqrt{3+h} - \sqrt{3+1}}{h}[/tex]

No, for two reasons (1) you forgot the "+1" in the first term and (2) you forgot the limit !!
[tex]f'(x)= \lim_{h\rightarrow 0}\frac{\sqrt{3+h+1}-\sqrt{3+1}}{h}[/tex]

Simplifying gives:
[tex]= 1[/tex]
No, it does not! Even forgetting the "+1" [itex]\sqrt{3+h}- \sqrt{3+1}[/itex] is NOT equal to h and so the fraction is not equal to 1!


But another time I didn't use the first principles:

[tex]f(x) = \sqrt{x + 1}[/tex]

[tex]1/2(x+1)^-1/2[/tex]

[tex]\frac{1}{1/2.\sqrt{x+1}}[/tex]
No,
[tex]\frac{1}{2\sqrt{x+1}}[/tex]

[tex]\frac{1}{1/2.\sqrt{3+1}} = 1[/tex]

(?) - "?"
[tex]\frac{1}{2\sqrt{3+1}}= \frac{1}{4}[/tex]

I got the same answer using both methods (=1). Am I on the right track? What else do I need to show?


Thank you. :smile:
 

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