MHB Help With Limit: a=16, f=${x}^{.25}$

[lastochka]

Hello, I have this homework questions with answers. I got part (a) a=16, but part (b) f=${x}^{.25}$ I don't understand...
Here is the problem:
This limit represents the derivative of some function f at some number a. State this a and f
$\lim_{{h}\to{0}}$$\frac{\sqrt[4]{16+h}-2}{h}$
Part a) is obvious the answer is 16 just by looking at it.
Please, help me to understand why in part b) f=${x}^{.25}$
Thank you in advance!

 

[Nono713]

Hint:
$$f(x) = x^{0.25} = \sqrt[4]{x}$$
Now:
$$f(16) = 16^{0.25} = \sqrt[4]{16} = 2$$
And:
$$f(16 + h) = (16 + h)^{0.25} = \sqrt[4]{16 + h}$$
Does that make it clearer?

 

[lastochka]

Hint:
$$f(x) = x^{0.25} = \sqrt[4]{x}$$
Now:
$$f(16) = 16^{0.25} = \sqrt[4]{16} = 2$$
And:
$$f(16 + h) = (16 + h)^{0.25} = \sqrt[4]{16 + h}$$
Does that make it clearer?

Why I didn't see that before! (Giggle)
Thank you so much for your help!