What is the domain of the expression #sqrt(7x+35)#?

1 Answer
Jun 10, 2017

Answer:

Domain: From #-5# to infinity
#[-5, oo)#

Explanation:

The domain means the values of #x# that make the equation untrue. So, we need to find the values that #x# cannot equal.

For square root functions, #x# cannot be a negative number. #sqrt(-x)# would give us #isqrt(x)#, where #i# stands for imaginary number. We cannot represent #i# on graphs or within our domains. So, #x# must be larger than #0#.

Can it equal #0# though? Well, let's change the square root to an exponential: #sqrt0 = 0^(1/2)#. Now we have the "Zero Power Rule", which means #0#, raised to any power, equals one. Thus, #sqrt0=1#. Ad one is within our rule of "must be greater than 0"

So, #x# can never bring the equation to take a square root of a negative number. So let's see what it would take to make the equation equal zero, and make that the edge of our domain!

To find the value of #x# the makes the expression equal to zero, let's set the problem equal to #0# and solve for #x#:

#0= sqrt(7x+35)#

square both sides

#0^2 = cancelcolor(black)(sqrt(7x+35)^cancel(2)#

#0=7x+35#

subtract #35# on both sides

#-35=7x#

divide by #7# on both sides

#-35/7 = x#

#-5 = x#

So, if #x# equals #-5#, our expression becomes #sqrt0#. That is the limit of our domain. Any smaller numbers than #-5# would give us a square root of a negative number.