## Section 1.4 Additional Exercises

¶These problems require a comprehensive knowledge of the skills reviewed in this chapter. They are not in any particular order. A proficiency in these skills will help you a long way as your learn the calculus material in the following chapters.

###### Exercise 1.4.1.

Rationalize the denominator for each of the following expressions. That is, re-write the expression in such a way that no square roots appear in the denominator. Also, simplify your answers if possible.

\(\dfrac{1}{\sqrt{2}}\)

\(\dfrac{3h}{\sqrt{x+h+1}-\sqrt{x+1}}\)

\(\dfrac{\sqrt{2}}{2}\)

\(3(\sqrt{x+h+1}+\sqrt{x+1})\)

###### Exercise 1.4.2.

Solve the following equations.

\(2-5(x-3)=4-10x\)

\(2x^2-5x=3\)

\(x^2-x-3=0\)

\(x^2+x+3=0\)

\(\sqrt{x^2+9}=2x\)

\(-13/5\)

\(-1/2,3\)

\((1\pm\sqrt{13})/2\)

No real solutions

\(\sqrt{3}\)

###### Exercise 1.4.3.

By means of counter-examples, show why it is wrong to say that the following equations hold for all real numbers for which the expressions are defined.

\((x-2)^2=x^2-2^2\)

\(\dfrac{1}{x+h}=\dfrac{1}{x}+\dfrac{1}{h}\)

\(\sqrt{x^2+y^2}=x+y\)

Counter-examples may vary.

\(x=3\)

\(x=h=1\)

\(x=y=1\)

###### Exercise 1.4.4.

Find an equation of the line passing through the point \((-2,5)\) and parallel to the line \(x+3y-2=0\text{.}\)

\(x+3y-13=0\text{,}\) or equivalents such as \(y=-\frac{1}{3}x+\frac{13}{3}\)

We first put the equation of the given line into standard form:

Thus, the slope of this line is \(-\frac{1}{3}\text{.}\) The line we seek needs to be parallel to this line, and so it must be of the form

To find \(b\text{,}\) we use the fact that the line passes through the point \((-2,5)\text{.}\) That is, it satisfies

From this, we see that \(b = \frac{13}{3}\text{.}\) Hence, the equation of the line is

###### Exercise 1.4.5.

Solve \(\dfrac{x^2-1}{3x-1}\leq 1\text{.}\)

\((-\infty,0]\cup(\frac{1}{3},3]\)

###### Exercise 1.4.6.

Explain why the following expression never represents a real number (for any real number \(x\)): \(\sqrt{x-2}+\sqrt{1-x}\text{.}\)

It is impossible for both \(x-2\) and \(1-x\) to be non-negative for the same real number \(x\text{.}\)

The domain of \(\sqrt{x-2}\) is \(x \geq 2\text{.}\) The domain of \(\sqrt{1-x}\) is \(x \leq 1\text{.}\) Since there exist no \(x \in \mathbb{R}\) which lie in both domains, the expression \(\sqrt{x-2}+\sqrt{1-x}\) does not ever represent a real number.

In other words, the domain of \(\sqrt{x-2}+\sqrt{1-x}\) is the intersection of \(\{x \in \mathbb{R} | x \geq 2\}\) and \(\{x \in \mathbb{R} | x \leq 1\}\text{,}\) which is the empty set, \(\emptyset\text{.}\)

###### Exercise 1.4.7.

Simplify the expression \(\dfrac{\left[3(x+h)^2+4\right]-\left[3x^2+4\right]}{h}\) as much as possible.

\(6x+3h\)

###### Exercise 1.4.8.

Simplify the expression \(\dfrac{\frac{x+h}{2(x+h)-1}-\frac{x}{2x-1}}{h}\) as much as possible.

\(-1/\left[(2x+2h-1)(2x-1)\right]\)

###### Exercise 1.4.9.

Simplify the expression \(-\sin x(\cos x+3\sin x)-\cos x(-\sin x+3\cos x)\text{.}\)

\(-3\)

###### Exercise 1.4.10.

Solve the equation \(\cos x=\frac{\sqrt{3}}{2}\) on the interval \(0\leq x\leq 2\pi\text{.}\)

\(\pi /6\text{,}\) \(5\pi /6\)

###### Exercise 1.4.11.

Find an angle \(\theta\) such that \(0\leq\theta\leq\pi\) and \(\cos\theta=\cos\frac{38\pi}{5}\text{.}\)

\(2\pi/5\)

###### Exercise 1.4.12.

What can you say about \(\dfrac{\left\vert x\right\vert+\left\vert 4-x\right\vert}{x-2}\) when \(x\) is a large (positive) number?

It is equal to 2 for all \(x\) larger than 4.

###### Exercise 1.4.13.

Find an equation of the circle with centre in \((-2,3)\) and passing through the point \((1,-1)\text{.}\)

\((x+2)^2+(y-3)^2=25\text{.}\)

###### Exercise 1.4.14.

Find the centre and radius of the circle described by \(x^2+y^2+6x-4y+12=3\text{.}\)

Centre is \((-3,2)\) and radius is 2.

###### Exercise 1.4.15.

If \(y=9x^2+6x+7\text{,}\) find all possible values of \(y\text{.}\)

\(y\) could be any real number greater than or equal to 6.

###### Exercise 1.4.16.

Simplify \(\left(\dfrac{3x^2 y^3 z^{-1}}{18x^{-1}yz^3}\right)^2\text{.}\)

\(x^6 y^4/(36z^8)\)

###### Exercise 1.4.17.

If \(y=\dfrac{3x+2}{1-4x}\text{,}\) then what is \(x\) in terms of \(y\text{?}\)

\(x=(y-2)/(3+4y)\)

###### Exercise 1.4.18.

Divide \(x^2+3x-5\) by \(x+2\) to obtain the quotient and the remainder. Equivalently, find polynomial \(Q(x)\) and constant \(R\) such that

\(Q(x)=x+1\text{,}\) \(R=-7\)