# domain of derivative

• Oct 25th 2006, 01:34 PM
gracy
domain 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 PM
Soroban
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 PM
ThePerfectHacker
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.