1. ## Combinations of functions

The question:

Given that,
$f(x) = x + \frac{1}{\sqrt{x}}$
$g(x) = \sqrt{x} - 2\sqrt{x}$

Find 6f + 3g.

My attempt:

$6(x + \frac{1}{\sqrt{x}}) + 3(\sqrt{x} - 2\sqrt{x})$
$6x + \frac{6}{\sqrt{x}} + 3\sqrt{x} - 6\sqrt{x}$

$6x + 3\sqrt{x}$

I'm not sure how they were able to simplify the expression that way. I'm sure it's something obvious that I'm overlooking. Any assistance would be great.

2. Can you double check the original problem? I'm curious as to why g(x) was written as
$g(x) = \sqrt{x} - 2\sqrt{x}$
when you could combine like terms and rewrite g(x) as
$g(x) = -\sqrt{x}$.

3. That's just how the textbook wrote it.

4. There is maybe a mistake in the textbook since, as already mentioned, you can simplify the expression of g by $\displaystyle g(x) = \sqrt{x} - 2\sqrt{x} = -\sqrt{x}$ and since the answer is not the one you get

I suggest that $\displaystyle g(x) = \sqrt{x} - \frac{2}{\sqrt{x}}$

In this case $\displaystyle 6f + 3 g = 6\left(x + \frac{1}{\sqrt{x}}\right) + 3\left(\sqrt{x} - \frac{2}{\sqrt{x}}\right) = 6x + 3\sqrt{x}$

5. Thanks. I guess I should check if there's errata for this text.