Question

find the domain of the function f(x)= 1/sqrt (4x-8)

please show calculations

Answer 1

huh

Answer 2

callum can you show the steps please

Answer 3

(2, infinity)

Answer 4

\[f(x) = \frac{1}{\sqrt{4x-8}}\] now you want to look at conditions on 4x-8. What can x reasonably be? Well you cannot take a square root of a negative number, and the denominator cannot be zero, so we need:
\[4x-8 > 0\]
\[x>\frac{1}{2}\]
So I was wrong above, the domain is actually:
\[\left(\frac{1}{2}, \infty \right)\]

Answer 5

x > 2

Answer 6

think you divided the wrong side there

Answer 7

oh yes sorry it's 2 lol, quite right.
\[(2,\infty)\]

Answer 8

rickjbr can you demonstrate how you got your answer please

Answer 9

you'll have imaginary numbers if the square root is allowed to go negative

Answer 10

thus not part of the real number system

Answer 11

callum was right in his calculation
4x - 8 > 0
4(x-2) > 0
x-2 > 0
x > 2

Answer 12

yeah my reasoning was right, it's just I divided the inequality wrong and got x>1/2. rickjbr corrected this to x>2

Answer 13

thanks for the help to both of your

Answer 14

np

Answer 15

how about this. 2x^2-3x+5...find and simplify the difference quotient

Answer 16

The question reads

Find and simplify the difference quotient f(x+h)-f(x)/h for the given function f(x)=2x^(2)-3x+5

Answer 17

ah, right

Answer 18

(2(x+h)^2-3(x+h)+5-2x^2+3x-5)/h

Answer 19

(2(x^2+2xh+h^2)-3(x+h)+5-2x^2+3x-5)/h

Answer 20

(2x^2+4xh+2h^2-3x-3h+5-2x^2+3x-5)/h

Answer 21

(2h^2+4xh+3x)/h

Answer 22

is that the final answer

Answer 23

checking, a bit rusty

Answer 24

my options are a)4x+2h, b)4x+2h-3, c)4h+2x-3, d)4h+2x, e)4x+4h

Answer 25

i'll redo it, might've missed something

Answer 26

ok

Answer 27

ah, silly

Answer 28

that last step was it. instead of (2h^2+4xh+3x)/h
it should be (2h^2+4xh-3h)/h

Answer 29

simply factor out an h

Answer 30

[h(2h+4x-3)]/h
=2h+4x-3

Answer 31

that's it

Answer 32

thanks greatly. i dont know if you have the time...but im reviewing for a test and I have a whole lot of other questions. if you can't help me, i understand, i will just post on the main forum

Answer 33

probably better off posting on main forum, if i can help, i'll try

Answer 34

ok thanks for the help

Answer 35

Find \[\frac{f(x+h)-f(x)}{h}\] for \[f(x)=2x^2-3x+5\]
\[\frac{f(x+h)-f(x)}{h} = \frac{(2(x+h)^2-3(x+h)+5) - (2x^2-3x+5)}{h}\]
\[= \frac{(2x^2+4xh+2h^2-3x-3h)-(2x^2-3x)}{h}\]
\[=\frac{4xh+2h^2-3h}{h} = 4x+2h-3\]

We can verify that this is right by noting that
\[f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h) - f(x)}{h}\]
The left hand side of this is equal to the derivative of f, which we can just do right here:
\[f'(x) = 4x - 3\]
and we should get that this equals the limit as h goes to zero of \[\frac{f(x+h) - f(x)}{h}.\] We have already calculated that \[\frac{f(x+h) - f(x)}{h} = 4x+2h-3\] so taking the limit of this as h goes to zero we get:
\[\lim_{h\rightarrow 0}(4x+2h - 3) = 4x+0-3 = 4x-3.\]
Hence we ave verified that the simplification has been done correctly.