# Question #210f3

Nov 6, 2016

5. (c)

6. The question is in error. The $3$ should be $4$ and the answer (d)

#### Explanation:

Question 5

$v \left(x\right) = \sqrt{\frac{x}{1 - x}}$

Note that if $x = 1$ then the denominator of $\frac{x}{1 - x}$ will be zero, so the quotient is undefined. So $1$ cannot be part of the domain.

That allows us to immediately eliminate options (d) and (e).

Note that for $x > 1$ the numerator will be positive and the denominator negative so $\frac{x}{1 - x} < 0$ and the square root is undefined (at least not Real).

So we can eliminate options (a) and (b).

That only leaves option (c), which is correct:

When $x < 0$, we find that $\frac{x}{1 - x} < 0$.

When $x \in \left[0 , 1\right)$ then $x \ge 0$ and $1 - x > 0$ so $\frac{x}{1 - x} \ge 0$ and the square root is Real valued.

So ${D}_{v} = \left[0 , 1\right)$

Let:

$y = v \left(x\right) = \sqrt{\frac{x}{1 - x}}$

Note that by definition of $\sqrt{\ldots}$ we must have $y \ge 0$

Then:

${y}^{2} = \frac{x}{1 - x} = \frac{1 - \left(1 - x\right)}{1 - x} = \frac{1}{1 - x} - 1$

Add $1$ to both ends to get:

${y}^{2} + 1 = \frac{1}{1 - x}$

Take the reciprocal of both sides to get:

$\frac{1}{{y}^{2} + 1} = 1 - x$

Add $x - \frac{1}{{y}^{2} + 1}$ to both sides to get:

$x = 1 - \frac{1}{{y}^{2} + 1}$

So for any $y \ge 0$ there is a value of $x$ such that $v \left(x\right) = y$

Hence the range of $v \left(x\right)$ is $\left[0 , \infty\right)$

$\textcolor{w h i t e}{}$
Question 6

$f \left(x\right) = \sqrt{4 - x}$

$g \left(x\right) = \sqrt{x + 3}$

Then:

${D}_{f} = \left(- \infty , 4\right]$

${D}_{g} = \left[- 3 , \infty\right)$

So:

${D}_{f + g} = {D}_{f} \cap {D}_{g} = \left(- \infty , 4\right] \cap \left[- 3 , \infty\right) = \left[- 3 , 4\right]$

This does not correspond to any of the given options, so there is an error in the question.

graph{sqrt(4-x) + sqrt(x+3) [-4.438, 5.56, -0.64, 4.36]}

$\left(f + g\right) \left(x\right)$ has minima at the ends of the domain and maximum in the centre.

${R}_{f + g} = \left[\sqrt{7} , \sqrt{14}\right]$

It looks like the error in the question was the $3$ instead of a $4$ and the correct answer would have been (d)