How do you find the domain of #f(x)=sqrt(1/2x-10)+3#?

2 Answers
Nov 24, 2016

Answer:

The domain is #D_f(x) in [20, +oo[ #

Explanation:

The domain is determined of what's under the square root sign.

#1/2x-10>=0#

#1/2x>=10#

#x>=20#

Therefore the domain of #f(x)# is #D_f(x) in [20, +oo[ #

Nov 24, 2016

Answer:

#Dom f = [20, + oo)#

Explanation:

The domain of a real function of real variable is the subset of the real numbers formed by those numbers for which the image exists:

#f : RR rarr RR#
#x in Dom f hArr f (x) in RR#

This assumes, in order to obtain the domain of the function, that when a function's formula only contains internal operations in #RR#, the domain of the function will be #RR# itself. In contrast, if the function formula uses non-internal operations in #RR#, for which values of the variable such operations give us a result that belongs to #RR#.

The square root is a non-internal operation in #RR# because we will only obtain a result that belongs to #RR# if the root is positive or zero. Roots of negative numbers generate imaginary numbers, which are not real.

So to know when the radicand is positive or negative, what we do is equate to zero:

#f (x) = root(n) {g (x)} , (n = dot 2) color(white) "..." rarr color(white) "..." g (x) = 0#,

and then we can define the domain of #f# as the set:

#Dom f = {x in RR | g (x) >= 0 }#.

To solve the domain of the function above:

#f (x) = sqrt {1/2 x - 10} + 3#

we must equal zero the term inside the root:

#1/2 x - 10 = 0#

and solve the equation obtained:

#x = 20#.

It is easy to see that if #x < 20# then #1/2 x - 10 < 0# that it means we have a non-real result. Consequently, the domain of the function will be the other values of #x#, that is:

#Dom f = {x in RR | x >= 20} = [20, + oo)#