- #1
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.
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.
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