Solving First Principles: Evaluating ƒ'(3)

[roam]

[SOLVED] The First Principles

Hello!
This is my question:

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



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( f'(x) = lim h\rightarrow0 \frac{f(x+h) - f(x)}{h} ):
∴ \frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}

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

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

Simplifying gives:
= 1


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

f(x) = \sqrt{x + 1}

1/2(x+1)^-1/2

\frac{1}{1/2.\sqrt{x+1}}

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

(?) - "?"

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


Thank you.

 

[Tedjn]

Please recheck what you get for this:

f'(x) = \frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}

Remember you need to substitute x+h for x.

 

[roam]

You mean like:
\frac{(x+h) - \sqrt{x+1}}{h}

\frac{(3+h) - \sqrt{3+1}}{h}

3 - 2 , → = 1

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?

 

[Gib Z]

No that's not what he meant :( Recheck your working, if f(x) = sqrt( x+1), what is f(x+h) ? Its not sqrt(x+h).

 

[HallsofIvy]

Hello!
This is my question:

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



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( f'(x) = lim h\rightarrow0 \frac{f(x+h) - f(x)}{h} ):
∴ \frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}

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

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

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

Simplifying gives:
= 1

No, it does not! Even forgetting the "+1" \sqrt{3+h}- \sqrt{3+1} is NOT equal to h and so the fraction is not equal to 1!

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

f(x) = \sqrt{x + 1}

1/2(x+1)^-1/2

\frac{1}{1/2.\sqrt{x+1}}

No,
\frac{1}{2\sqrt{x+1}}

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

(?) - "?"

\frac{1}{2\sqrt{3+1}}= \frac{1}{4}

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


Thank you.