Let f be a function defined by: ?

Let f be a function defined by:

State the domain of "f" and find the value(s) of a for which "f" is a continuous function.

Answer 1

The Domain of the Function is:
#x in [-2/7,oo)#
or
#x ≥ -2/7#

The value of a is #1/8#.

So the first thing the question is asking is "What is the domain of the function?" The domain of a function is the list of #x# values which return something. For example, plugging in 0 to the function returns 0.2929, meaning 0 is in the domain. Plugging in -5 would return the square root of a negative, which we'll be ignoring here.
The piecewise physically limits our domain, but that's not actually the domain, since we have a square root with an x in it. Since we're going to be ignoring imaginary numbers (otherwise the doman would be #x in RR#), we have to solve these. The one to reach the negative values faster will limit our domain.
In this case, #sqrt(7x+2)# reaches 0 faster. Solving the inside tells us that #x# has to be greater than or equal to #-2/7# in order to return a value. Hence, our domain is limited to
#x in [-2/7,oo)# or #x ≥ -2/7#
The next question asks "Find the value of #a# which makes #f(x)# a continuous function." It's important to understand what continuity means. Continuous functions are functions which have no break, or leap, in their value. If we were only to graph that rational function, it would result in a hold at #x=2# because we would be dividing by 0. Therefore, the function is not continuous at that point. What we need is to find what that value would be if we weren't dividing by 0.

So how do we do that? Well we have different ways of going about this. The first, and simplest, is to graph the function and see what happens at the value.

graph{(sqrt(7x+2)-sqrt(6x+4))/(x-2) [1.6815, 2.2657, -0.0507, 0.2413]}

As we can see, the function clearly approaches #1/8#. But what if we can't use the calculator? Well we need to be able to manipulate the function until we get something we want.
#(sqrt(7x+2)-sqrt(6x+4))/(x-2)#

I want to rationalize the numerator, so that I get rid of the square roots up there. So I'll multiply by the conjugate.

#(sqrt(7x+2)-sqrt(6x+4))/(x-2)*((sqrt(7x+2)+sqrt(6x+4))/(sqrt(7x+2)+sqrt(6x+4)))#

Now, let's simplify it.

#(7x+2-6x-4)/((x-2)(sqrt(7x+2)-sqrt(6x+4))#

Now, we can combine like terms and simplify even further.

#(x-2)/((x-2)(sqrt(7x+2)-sqrt(6x+4))#
This is perfect! Now we can cancel out the x-2 on the top and bottom, thus eliminating the hole at #x=2#
#1/(sqrt(7x+2)-sqrt(6x+4))#

Now, just plug in 2, and get:

#a = 1/8#
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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