Question

can someone show me step by step on how to do the attach problem?

Answer 1

i take it thats a 4rt(..) ?

Answer 2

its hard to jump back and forth with these things; itd be better if you could learn to type them out at least :)

Answer 3

usually you multiply by a conjugate form

Answer 4

\[\lim_{h->0}\frac{\sqrt[4]{16+h}-2}{h}\]

Answer 5

yeah, but its take me a long time to do that

Answer 6

its spose to, that teaches you the math discipline :)

Answer 7

True!

Answer 8

isnt this number 2:- f(x) = x^(1/4)

Answer 9

But, again when I try to type euqation its doesn't always come out clear enough

Answer 10

on this website

Answer 11

\[\lim_{h->0}\frac{\sqrt[4]{16+h}-2}{h}\]
\[\lim_{h->0}\frac{\sqrt[4]{16+h}-2}{h}*\frac{\sqrt[4]{16+h}+2}{\sqrt[4]{16+h}+2}\]
\[\lim_{h->0}\frac{\sqrt[4]{(16+h)^2}-4}{h\sqrt[4]{16+h}+2}\]
that could have worked out better :) conjugate again?

Answer 12

isn't the square root on the numerator suppose to cancel out?

Answer 13

yeah .... lets try that way too :)

Answer 14

\[\lim_{h->0}\frac{\sqrt{(16+h)}-4}{h\sqrt[4]{16+h}+2}*\frac{\sqrt{(16+h)}+4}{\sqrt{(16+h)}+4}\]
\[\lim_{h->0}\frac{16+h-4}{h(\sqrt[4]{16+h}+2)(\sqrt{16+h}+4)}\]
\[\lim_{h->0}\frac{12+h}{h(\sqrt[4]{16+h}+2)(\sqrt{16+h}+4)}\]

Answer 15

this right so far?

Answer 16

yes!

Answer 17

you agreeing or verifying :)

Answer 18

both

Answer 19

wait! on the numerator its suppose to be 16+h-16

Answer 20

\[\lim_{h->0}\frac{12+h}{h((16+h)^{1/4}+2)((16+h)^{1/2}+4)}\]
\[\lim_{h->0}\frac{12+h}{h((16+h)^{3/4}+4(16+h)^{1/4}+2(16+h)^{1/2}+8)}\]

good eye :)

Answer 21

then lets cancel hs and solve for the others for 0

Answer 22

\[\lim_{h->0}\frac{h}{h((16+h)^{1/4}+2)((16+h)^{1/2}+4)}\]
\[\lim_{h->0}\frac{1}{((16+h)^{1/4}+2)((16+h)^{1/2}+4)}\]
\[\frac{1}{((16)^{1/4}+2)((16)^{1/2}+4)}\]
\[\frac{1}{(2+2)(4+4)}\]
\[\frac{1}{(4)(8)}=\frac{1}{32}\]
no?

Answer 23

That right but, its not one of the answer choices.

Answer 24

http://www.wolframalpha.com/input/?i=lim {h-%3E0}+%28%2816%2Bh%29^%281%2F4%29-2%29%2Fh

Answer 25

well have to consider what the problem is asking then

Answer 26

which function given is the value 1/32?

Answer 27

the problem asks you to find the function which leads to this limit and also the value of a

Answer 28

youll have to derive the given functions and relate them at f'(a) i think right?

Answer 29

right

Answer 30

i think its number 2 : x^1/4 a = 16

Answer 31

can you please show how you got that answer?

Answer 32

i think number 2 as well;

f'(x) = \(\cfrac{1}{\sqrt[4]{x^3}}\) ; at x=16?

Answer 33

???

Answer 34

that give 1/8

Answer 35

ok

Answer 36

unless we did an error in the first part; number 2 looks the best fit

Answer 37

and by we i meant me lol

Answer 38

if you work back and differentiate x^1/4 from first principles it leads to the limit quoted in the question - according to my workings

Answer 39

what about the answer amistre64 got 1/32, was that necessary?

Answer 40

.... i hoe so, that was alot of work :)

Answer 41

pppp hope

Answer 42

so why do you think the answer is choice 2

Answer 43

you got more than 6 answers to choose from?

Answer 44

wolfram says 1/32 was right if the link hadnt chopped

Answer 45

no, just 6

Answer 46

i forgot the 4 .... doh!

Answer 47

\[[x^{1/4}]'=\frac{1}{4x^{3/4}}\]

Answer 48

thats why it told me 8, i forgot the 4 :)

1/8(4) = 1/32

Answer 49

oh, ok! also the answer choice has a=16, why?

Answer 50

they use "a" instead of "x" to indicate that its a specific number in the domain

Answer 51

f'(a) such that a=16

Answer 52

also where did you get X^3 from they only gave X^4

Answer 53

instead of saying; go get me one of those pencils over there. They are saying: go get me that pencil over there

Answer 54

y = x^(1/4)
if y increases by a quantity dy and x increases by dx
y + dy = (x + dx)^4
dy = (x + dx)^1/4 - x^ 1/4
dy/dx = [(x + dx)^1/4 - x^ 1/4 ]/ dx

for dx insert h
and for x insert 16
limit x>0 = [(16 - h)^1/4 - 16^1/4] / h
= [(16 - h)^1/4 - 2 / h

Answer 55

we need a function that goes under; and fractional exponents do that since you subtract 1 from it and it goes negative

Answer 56

\[\frac{d}{dx}{x^{1/4}} \implies\frac{1}{4}\frac{1}{x^{3/4}} \]

Answer 57

when x = a = 16; we get a value that is 1/32

Answer 58

oh, ok thanks some much!

Answer 59

youre welcome

Answer 60

how come you had to take the conjugate two time?

Answer 61

howdy

Answer 62

the conjunction only resolves a square root; this was a 4th root

Answer 63

as a result, the first conjugate canceled out a square and left us with a remaining square to deal with

Answer 64

in other words:
\[x^{1/4}*x^{1/2}=x^{1/2}\]
\[x^{1/2}*x^{1/2}=x\]

Answer 65

oh, ok thanks for clearing it up for me

Answer 66

when you get a new question, feel free to post it to the left so others can see it as well and help out :)