Question

for f(x)=√2x-1, find the following
a. the domain of f(x)
b. f(2)
c. f(x+h)

Answer 1

hi,
for the first part
is it
\[\sqrt{2x -1} \ or \ \sqrt{2x} -1\]
I think it should be the first one ??

Answer 2

no i need help with like can you find f(x) also find f(2) and find f(x+h)

Answer 3

yeah !! I can help u , But I need to be sure about the expression first...
the expression you've typed .. is it
\[f(x) = \sqrt{2x -1}\]
or
\[f(x) = \sqrt{2x} -1\]

the way u have use square root sign, it's not clear !!!

Answer 4

oh the first one where it covers the whole problem

Answer 5

got it...
and ya want to find f(2) first....

to find f(2) , substitute 2 instead of "x" in f(x) ... i.e -\[f(x) = \sqrt{2x -1}\]\[f(2) = \sqrt{(2\times2) -1}\]

now what is the value of f(2) ?

Answer 6

i'm really bad at math im not sure =(

Answer 7

nuh.... just forget it...
what is the value of this one..
(2x2) -1 ?

Answer 8

f(2)?

Answer 9

\[f(2) = \sqrt{(2 \times 2 ) -1}\]\[f(2) = \sqrt{4 -1}\]\[f(2) = \sqrt{3}\]

Did ya get it now...?
Shall we move to the next part ?

Answer 10

yes

Answer 11

so....
the next one asks for f(x + )
to find f(x +h )
u have to substitute "(x + h)" instead of "x" in f(x)
i.e
\[f(x) = \sqrt{2x -1}\]\[f(x + h ) = \sqrt{2(x + h) - 1}\]\[f(x + h ) = \sqrt{2x +2h- 1}\]

that's it !!! did ya get it ?

Answer 12

yes it's making sense what's the domain of f(x)

Answer 13

do u know the domain of f(x) ?

Answer 14

no i was hopin you could help me with that

Answer 15

do u know it?

Answer 16

would it be x≥1/2

Answer 17

domain of f(x) means the all possible values that u can use instead of "x" in f(x)..
i.e
2 is a number in the domain of f(x)...
because u substitute 2 and u will get a rational answer...

but what about -1 ?
lets see what will happen to f(-1)
\[f(-1) = \sqrt{[2 \times (-1 ) ] -1} \]
\[f(-1) = \sqrt{-2-1}\]\[f(-1) =\sqrt{-3}\]

Ops!! we've got a negative square root as the answer.. as negative square roots can't be defined rationally .... -1 doesn't belong to the domain of f(x)

So.. u have to find the range of value that will give u a positive square root for f(x)...
So.. the minimum value for f(x) is 0 ... any value below that will be a negative square root...
U will get 0 for f(x) when u find f(1/2)...
i.e - when the value of "x" is 1/2 the value of f(x) is 0
the minimum value that f(x) can take is 0
the value of "x" that makes f(x) = 0 is 1/2
so .... the domain of f(x) is
\[\left\{ x/x \ \epsilon \ \mathbb{R} ; x \ge 1/2 \right\}\]