Precalculus: Lecture Notes

By definition of the function \(f(x) = \lceil x \rceil\text{:}\)

We need to find all real numbers \(x\) with the property that \(f(x) = \lceil x \rceil =1\text{.}\) In other words, we are looking for all real numbers such that the number 1 is the smallest integer that is greater than or equal to \(x\text{.}\)

This gives the set

See Figure 5.14.

We note that there is a vertical line that intersects Curve A at two points. See Figure 5.17.

This means, by the vertical line test, that Curve A does not represent a graph of a function. We see from Figure 5.18 that every vertical line intersects Curve B in at most one point. Hence, Curve B represents a graph of a function. Therefore:

1. From the graph of \(f\) (Curve \(B\)) we read that the function \(f\) is defined on the interval \([-2,2)\text{.}\) The range of the function is the interval \((-4,3]\text{.}\)

2. We read from the graph:

   \begin{equation*} \begin{array}{c||c|c|c|c} x\amp -2\amp -1\amp 0\amp 3\\ \hline f(x)\amp 3\amp 0\amp 0\amp\text{Not defined}\\ \end{array} \end{equation*}

3. Based on the graph it appears that \(f(-0.75)\approx 0.75\) and \(f(1.99999)\approx -4\text{.}\)

4. From the graph we read that the equation \(f(x)=0\) has three solutions: \(-1\text{,}\) \(0\text{,}\) and \(1\text{.}\)

To complete the table we use the given definition of the function \(f\) for \(x\in [0,2]\) and the fact that for odd functions we have \(f(-x)=-f(x)\) for all \(x\) in the domain of \(f\text{.}\) For example, since, from the above definition, \(f(2)=2\text{,}\) it follows that \(f(-2)=-f(2)=-2\text{.}\)

First we draw the graph of \(f\) for \(x\in [0,2]\) according to the given definition. Then we use the fact that \(f\)is an odd function and that its graph must be symmetric with the respect to the origin. See Figure 5.19.

See Figure 5.20 – Figure 5.23.

1. Note that the function \(g\) is not defined for \(x=1\text{.}\) It follows that for all \(x\not= 1\text{:}\)

   \begin{equation*} \begin{array}{rcl} (f\circ g)(x)\amp=\amp f(g(x))=f\left( \frac{x+1}{x-1}\right)\\ \amp=\amp \frac{x+1}{x-1}+\frac{1}{\frac{x+1}{x-1}}=\frac{x+1}{x-1}+\frac{x-1}{x+1}\\ \amp=\amp \frac{(x+1)^2+(x-1)^2}{(x+1)(x-1)}=\frac{x^2+2x+1+x^2-2x+1}{x^2-1}\\ \amp=\amp \frac{2x^2+2}{x^2-1}=\frac{2(x^2+1)}{x^2-1}.\\ \end{array} \end{equation*}

   Hence the domain of the function \(f\circ g\) is the set \(\{x:x\not= 1 \text{ and }x^2-1\not=0\}=\{x:x\not= 1 \text{ and }(x+1)(x-1)\not=0\}=\{ x: x\not= 1, x\not= -1\}\text{.}\)

2. Note that the function \(f\) is not defined for \(x=0\text{.}\) It follows that for all \(x\not= 0\text{:}\)

   \begin{equation*} \begin{array}{rcl} (g\circ f)(x)\amp =\amp g(f(x))=g\left( x+\frac{1}{x}\right)\\ \amp =\amp \frac{x+\frac{1}{x}+1}{x+\frac{1}{x}-1}=\frac{\frac{x^2+1+x}{x}}{\frac{x^2+1-x}{x}}\\ \amp =\amp \frac{x^2+x+1}{x^2-x+1}\\ \end{array} \end{equation*}

   We observe that, for all real numbers \(x\text{,}\)

   \begin{equation*} x^2-x+1=x^2-2\cdot \frac{1}{2}\cdot x+\frac{1}{4}+\frac{3}{4}=\left( x-\frac{1}{2}\right)^2 +\frac{3}{4}>0. \end{equation*}

   Hence the domain of the function \(g\circ f\) is the set \(\{ x: x\not= 0\}\text{.}\)

3. We observe that the function \(f\circ g\) is defined for \(x=0\) and that the \(g\circ f\) is not. Hence those two functions are not equal.

We can take

Then

We note that there is a horizontal line that intersects the graph of Function B at two points. This means, by the horizontal line test, that Function B is not a one–to–one function. We \textit{see} that every horizontal line intersects the graph of Function A at most one point. Hence, Function A is a one–to–one.

See Figure 5.26 and Figure 5.27.

Therfore:

1. See Figure 5.28.

   

2. Complete the following table and fill in all boxes:

   \begin{equation*} \begin{array}{ccl} \text{Domain of }f\amp=\amp [-2,1]\\ \hline \text{Range of }f\amp = \amp [-1,2]\\ \hline \text{Domain of }f^{-1}\amp = \amp [-1,2]\\ \hline \text{Range of }f^{-1} \amp = \amp [-2,1]\\ \hline (f\circ f^{-1})(x) \amp = \amp x , \text{for all }x\in [-1,2] \\ \hline (f^{-1}\circ f)(x)\amp = \amp x, \text{ for all }x\in [-2,1] \\ \hline ((f^{-1})^{-1})(x) \amp = \amp f(x), \text{ for all }x\in [-2,1] \\ \end{array} \end{equation*}

This is a linear equation. We note that the given equation is equivalent to the equation

From

we conclude that the given equation has no solution.

The given equation is a quadratic equation. From

it follows that \(x=\frac{1}{2}\) is the only equation of the given equation.

We note that the substitution \(t=5^x\) gives a quadratic equation \(t^2-t-12=0\text{.}\)

The quadratic formula implies

Since \(t=5^x>0\text{,}\) we reject the negative solution of the quadratic equation above, \(t=\frac{1-7}{2}=-3\text{.}\)

Hence

From

we conclude that all points on the line \(/ell\) are of the form \((x,-2x+1)\text{,}\) \(x\in \mathbb{R}\text{.}\)

It follows that the distance between a point \((x,-2x+1)\) on the line and the point \((-1,1)\) is given by

The quadratic function \(f(x)=5x^2+2x+1\) achieves its minimum at the first coordinate of the vertex of its graph:

The point on the line\(\ell\)} that is closest to the point \(P\) is

See Figure 5.29.

The surface area is the sum of the areas of the square that is the bottom of the box and the areas of the four sides. Note that each side is a rectangle with sides \(x\) and \(h\text{.}\) Hence the surface area is given by \(S=x^2+4hx\text{.}\)

Next we express the variable \(h\) in terms of the variable \(x\text{.}\)

The volume of the box is given as the product of the area of the base and the hight of the box. Since the base is a square with the side \(x\) we have that \(V=x^2h\text{.}\)

It is given that \(V=1\) so

It follows that

Observe that the expression for \(S\) obtained above is defined for all \(x\not= 0\text{.}\) Since \(x\) is the length of the side of the base, we have that \(x\) must be a positive real number.

Thus, the domain of the function \(S\) is the set \(\{x\in \mathbb{R}|x>0\}=(0,\infty)\text{.}\)

If \(x\) is a large number then the term \(\frac{4}{x}\) is very small and \(S\approx x^2\text{.}\) This means that the box will have a large base but that the hight of the box will be small.

If \(x\) is a small number then the term \(x^2\) is very small and \(S\approx \frac{4}{x}\text{.}\) This means that the box will have a large hight but that the base of the box will be small.

Since \(f\) is a rational function, its domain is the set of all real numbers for which the denominator of \(f\) is not equal to 0.

We solve the equation \(x^2-x-2=0\) by using the quadratic formula (or any other method):

Therefore, the domain of the function \(f\) is the set \(\{x\in \mathbb{R}: x\not= -1, x\not= 2\}\)

. Note that

We observe that \(x^2\geq 0\) for all real numbers. Thus, \(x^2+2\gt 0\) for all real numbers which implies that the numerator of \(f\) is never equal to 0.

This means that \(f\) has no zeros.

The function \(f\) is not defined at \(x=-1\) and \(x=2\text{.}\) Since the numerator of \(f\) does not cancel at these values of \(x\) , we conclude that the vertical asymptotes are lines \(x=-1\) and \(x=2\text{.}\)

The \(y\)–intercept of \(f\) is given by \(f(0)=\frac{0^2+1}{0^2-0-2}=-\frac{1}{2}\text{.}\)

We use the fact that for all \(x\) with large absolute values a polynomial in \(x\) behaves as its leading term. Thus

This means that the line \(y=1\) is a horizontal asymptote to the graph of \(f\text{.}\)

From \(f(x)=\frac{x^2+1}{x^2-x-2}\) and the fact that \(x^2+1>0\) for all real number \(x\) we conclude that

The graph of the parabola \(y=x^2-x-2=(x+1)(x-2)\text{,}\) a “happy” parabola with zeros at \(x=-1\) and \(x=2\text{,}\) gives that

Therefore

The domain of the function \(f\) is given by

We solve the equation \(f(x)=2\log(x+1)+2=0\text{.}\)

It follows that

Hence \(x=-1+10^{-1}=-0.9\) is the only zero of the function \(f\text{.}\)

To find \(f^{-1}\) we start with \(y=2\log(x+1)+2\text{,}\) interchange the roles of “\(y\)”and “\(x\)” and then solve

for \(y\text{.}\)

It follows that

Hence

See Figure 5.31 – Figure 5.34.