What would the instantaneous rate of f(x) = √x + 3 be?

Question

mathstudent55 · Answer

Is it 
$y = \sqrt{x} + 3$ 
or
$y = \sqrt{x + 3} $   ?

anonymous · Answer

the first one

anonymous · Answer

would it be 1/6?

mathstudent55 · Answer

$f(x) = \sqrt{x} + 3$

You need to take the derivative of y with respect to x.

$f'(x) = \dfrac{d}{dx}(\sqrt{x} + 3)$

$f'(x) = \dfrac{d}{dx}\sqrt{x} + \dfrac{d}{dx}3$

$f'(x) = \dfrac{d}{dx}x^{\frac{1}{2}} + 0$

$f'(x) = \dfrac{1}{2} x^{-\frac{1}{2}} + 0$

$f'(x) = \dfrac{1}{2x^{\frac{1}{2}}}$

$f'(x) = \dfrac{1}{2\sqrt{x}}$

Now if you are told a specific x value at which you need the rate of change, plug into 

$f'(x) = \dfrac{1}{2\sqrt{x}}$

anonymous · Answer

I haven't learned derivatives yet gaaaah

mathstudent55 · Answer

Then what have you learned already that allows you to find the instataneous rate of change?

anonymous · Answer

Average R.O.C = f(x) - f(x)/x 
And then we assume that x is pretty much 0 at the end because it will be so close to 0.. Just learned it today..

anonymous · Answer

Should I take a picture of the note?

mathstudent55 · Answer

Are you asked a specific x value to find the r.o.c. for?

anonymous · Answer

Yes

anonymous · Answer

and assume x is 0 because it is pretty much 0..

anonymous · Answer

Does it help if i tell you that we also use it to find slope of tangent?

mathstudent55 · Answer

I just don't know this way of doing it, sorry.

anonymous · Answer

No problem :) Thanks a bunch :)

agent0smith · Answer

$$\Large \lim_{h \rightarrow 0} \frac{ f(a+h) - f(a) }{ h }$$is it something like that?