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2. ## re: A few questions involving square roots and inequalities

For the first one, $\displaystyle \sqrt{x - 1}$ is not equal to $\displaystyle \sqrt{x} - 1$.

You should start by squaring both sides.

3. ## re: A few questions involving square roots and inequalities

Originally Posted by juliak

For the last one, you first need to note that $\displaystyle x \neq 3$ (Why?)

Then

$\displaystyle \begin{align*} \frac{\sqrt{x^3 - 6x^2 + 9x}}{x-3} &= \frac{\sqrt{x(x^2 - 6x+ 9)}}{x - 3} \\ &= \frac{\sqrt{x}\sqrt{x^2 - 6x + 9}}{x - 3} \\ &= \frac{\sqrt{x}\sqrt{(x - 3)^2}}{x-3} \textrm{ which means }x \geq 0 \\ &= \frac{\sqrt{x}|x-3|}{x-3} \\ &= \begin{cases}\frac{\sqrt{x}(x - 3)}{x-3}\textrm{ if }x - 3 > 0 \\ \frac{\sqrt{x}\,\left[-(x-3)\right]}{x-3}\textrm{ if }x - 3 < 0\end{cases} \\ &= \begin{cases}\sqrt{x}\textrm{ if }x > 3 \\ -\sqrt{x} \textrm{ if }0 \leq x < 3 \textrm{ since we have already established that }x \geq 0\end{cases}\end{align*}$

You should be able to sketch this from here.

4. ## re: A few questions involving square roots and inequalities

Originally Posted by juliak

b) Your working has been marked incorrect because you have failed to realise that multiplying or dividing both sides of an inequality changes the inequality sign. It's quite possible that your denominators could be negative. The easiest way to do this is first to note that $\displaystyle x \neq -1$ and $\displaystyle x \neq 1$, then...

$\displaystyle \begin{align*} \frac{1}{1 + x} &< \frac{x}{x - 1} \\ \frac{1}{1 + x} &< \frac{x - 1 + 1}{x - 1} \\ \frac{1}{1 + x} &< \frac{x - 1}{x - 1} + \frac{1}{x - 1} \\ \frac{1}{1 + x} &< 1 + \frac{1}{x - 1} \\ \frac{1}{1 + x} - \frac{1}{x - 1} &< 1 \\ \frac{(x - 1) - (1 + x)}{(1 + x)(x - 1)} &< 1 \\ \frac{-2}{(1 + x)(x - 1)} &< 1 \end{align*}$

Now to solve this, you need to consider two cases, the first where $\displaystyle (1 + x)(x - 1) < 0$ and the second where $\displaystyle (1 + x)(x - 1) > 0$.

Case 1:
$\displaystyle (1 + x)(x - 1) < 0 \implies x^2 - 1 < 0 \implies |x| < 1 \implies -1 < x < 1$
which gives

$\displaystyle \begin{align*} \frac{-2}{(1 + x)(x - 1)} &< 1 \\ -2 &> (1 + x)(x - 1) \\ -2 &> x^2 - 1 \\ x^2 - 1 &< -2 \\ x^2 &< -1 \textrm{ which is not true for any }x \end{align*}$

Case 2:
$\displaystyle (1 + x)(x - 1) > 0 \implies x^2 - 1 > 0 \implies |x| > 1 \implies x < -1 \textrm{ or }x > 1$
which gives

$\displaystyle \begin{align*} \frac{-2}{(1 + x)(x - 1)} &< 1 \\ -2 &< (1 + x)(x - 1) \\ -2 &< x^2 - 1 \\ x^2 - 1 &> -2 \\ x^2 &> -1\textrm{ which is true for all possible }x \end{align*}$

So the solution is $\displaystyle x < -1 \textrm{ and }x > 1$.

5. ## re: A few questions involving square roots and inequalities

Originally Posted by juliak

$\displaystyle \begin{align*} |4x - x^3| &= \begin{cases}4x - x^3 \textrm{ if }4x - x^3 \geq 0 \\ x^3 - 4x \textrm{ if }4x - x^4 < 0\end{cases} \end{align*}$

So to work out the possible $\displaystyle x$ values for each case, we need to solve the equation $\displaystyle f(x) = 4x - x^3 = 0$, since the function changes sign at these solutions.

$\displaystyle \begin{align*} 4x - x^3 &= 0 \\ x(4 - x^2) &= 0 \\ x(2 - x)(2 + x) &= 0 \\ x = -2 \textrm{ or }x &= 0 \textrm{ or }x = 2 \end{align*}$

Now testing values for the function...

$\displaystyle \begin{align*} f(-3) &= 4(-3) - (-3)^3 \\ &= -12 - (-27) \\ &= 27 - 12 \\ &= 15 \end{align*}$

so $\displaystyle 4x - x^3 > 0\textrm{ if } x < -2$.

$\displaystyle \begin{align*}f(-1) &= 4(-1) - (-1)^3 \\ &= -4 - (-1) \\ &= 1 - 4 \\ &= -3\end{align*}$

so $\displaystyle 4x - x^3 < 0 \textrm{ if }-2 < x < 0$.

$\displaystyle \begin{align*} f(1) &= 4(1) - 1^3 \\ &= 4 - 1 \\ &= 3 \end{align*}$

so $\displaystyle 4x - x^3 > 0 \textrm{ if } 0 < x < 2$.

$\displaystyle \begin{align*} f(3) &= 4(3) - 3^3 \\ &= 12 - 27 \\ &= -15 \end{align*}$

so $\displaystyle 4x - x^3 < 0 \textrm{ if } x > 2$.

Therefore

$\displaystyle |4x - x^3| = \begin{cases} 4x - x^3 \textrm{ if }x \leq 2 \textrm{ or }0 \leq x \leq 2 \\ x^3 - 4x \textrm{ if }-2 < x < 0 \textrm{ or }x > 2 \end{cases}$

6. ## Re: A few questions involving square roots and inequalities

$\sqrt(x - 1) = \sqrt(1-x)$ is true only if x = 1 which is the solution.

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