Domain of f(x)=sqrt(1+2x)

1+2x>=0

x>=-1/2

domain:: (-1/2,+inf)

and

domain of f '(x)=1/sqrt(1+2x).........do not understand to do this one..

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- Oct 25th 2006, 01:34 PMgracydomain of derivative
Domain of f(x)=sqrt(1+2x)

1+2x>=0

x>=-1/2

domain:: (-1/2,+inf)

and

domain of f '(x)=1/sqrt(1+2x).........do not understand to do this one.. - Oct 25th 2006, 02:08 PMSoroban
Hello, Gracy!

A minor (but important) change in your first answer . . .

Quote:

Domain of $\displaystyle f(x) = \sqrt{1+2x}$

$\displaystyle 1 + 2x \:\geq \:0\quad\Rightarrow\quad x\:\geq \:-\frac{1}{2}$

Domain: .$\displaystyle \left[-\frac{1}{2},\:\infty\right)$

. . . . . . . ↑

We__include__the $\displaystyle -\frac{1}{2}$

Quote:

Domain of $\displaystyle f'(x) \:=\:\frac{1}{\sqrt{1+2x}}$

That prime-mark is distracting, but we can ignore it.

Since $\displaystyle 1 + 2x$ is under a square root, it must be positive or 0.

But since it is in the denominator, it cannot equal 0.

Hence, we have: .$\displaystyle 1 + 2x\:>\:0\quad\Rightarrow\quad x \:>\:-\frac{1}{2}$

Domain: .$\displaystyle \left(-\frac{1}{2},\:\infty\right)$

- Oct 25th 2006, 04:34 PMThePerfectHacker
Basically whenever you have a continous function on [a,b] and differenciable at (a,b) then the derivative does not exist at 'a' and 'b'. Do you see why? One of the conditions that implies a function is not differenciable is when it fails to be countinous. Note whenever you have a continous function on [a,b] then it is continous from the right at 'a' and continous from the left at 'b' but it is not fully continous.