A few questions involving square roots and inequalities

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September 6th 2011, 11:08 PM
juliak

A few questions involving square roots and inequalities

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September 6th 2011, 11:49 PM
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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.
September 7th 2011, 12:04 AM
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re: A few questions involving square roots and inequalities

Quote:

Originally Posted by juliak

http://img841.imageshack.us/img841/4636/31225000.png

http://img221.imageshack.us/img221/162/13882076ij.png

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.
September 7th 2011, 12:20 AM
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re: A few questions involving square roots and inequalities

Quote:

Originally Posted by juliak

http://img841.imageshack.us/img841/4636/31225000.png

http://img221.imageshack.us/img221/162/13882076ij.png

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 $\displaystyle x < -1 \textrm{ and }x > 1$ .
September 7th 2011, 12:34 AM
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re: A few questions involving square roots and inequalities

Quote:

Originally Posted by juliak

http://img841.imageshack.us/img841/4636/31225000.png

http://img221.imageshack.us/img221/162/13882076ij.png

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

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

$\displaystyle |4x - x^3| = \begin{cases} 4x - x^3 \textrm{ if }x \leq 2 \textrm{ or }0 \leq x \leq 2 \\ x^3 - 4x \textrm{ if }-2 < x < 0 \textrm{ or }x > 2 \end{cases}$
September 7th 2011, 07:13 AM
mathjeet

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.