A kosarad üres
More examples of functions
(higher level courseware)
We got to know some basic functions and functions which can be derived from these with the help of some simple rewritings and transformations. Now we are dealing with further examples of some more complex functions.
Example 1
Let us examine and plot the graph of function \latex{f: (\R\backslash\left\{0\right\})\rightarrow\R}, \latex{f(x)=x+\frac{1}{x}}.
Solution
Function \latex{f} is odd, because if \latex{x\neq 0}, then
\latex{f(-x)=-x+\frac{1}{-x}=-\left(x+\frac{1}{x}\right)=-f(x)}.
Its graph is symmetric about the origin, thus it is enough to draw it at positive \latex{x} values, and then to reflect the resulting curve about the origin. For positive x, x and \latex{\frac{1}{x}} should be added. We know the images of both functions well; we could even “add” the graphs graphically point by point. (Figure 79)At small positive places \latex{x} the graph almost touches the image of \latex{x\mapsto\frac{1}{x}}, and at large positive places \latex{x} it almost touches straight line \latex{y = x}. It can graphically be seen that the function has a minimum at \latex{1}, its minimum value is \latex{2}. We are going to show it now.
\latex{y=x+\frac{1}{x}}
Figure 79
\latex{y=x}
\latex{y=\frac{1}{x}}
\latex{ 2 }
\latex{ 1 }
\latex{ 1 }
\latex{ y }
\latex{ x }
If \latex{a\gt 0}, then \latex{a+\frac{1}{a}\geq2}, and the equality is true only when \latex{a=1}.
Let us show that the difference of the left side and the right side is non-negative, and it is 0 only when \latex{a=1}.
Let us show that the difference of the left side and the right side is non-negative, and it is 0 only when \latex{a=1}.
\latex{a+\frac{1}{a}-2=\frac{a^2+1-2a}{a}=\frac{(a-1)^2}{a}\geq0},
since \latex{a\gt0} and \latex{(a-1)^2\geq0}, the equality is fulfilled only when \latex{a=1}.⯁ ⯁ ⯁
It is worth formulating the shown inequality, which can be used very often, also in words.
The sum of a positive number and its reciprocal is not less than \latex{ 2 }, and it is equal to \latex{ 2 } only for the number \latex{ 1 }.
If \latex{x\gt0}, then \latex{x+\frac{1}{x}\geq2}.
It is also easy to show that we indeed drew the curve correctly, at positive places \latex{x} function \latex{x\mapsto x+\frac{1}{x}} is convex.
Figure 80
\latex{P(a; a+\frac{1}{a})}
\latex{Q(b; b+\frac{1}{b})}
\latex{a}
\latex{b}
\latex{\frac{a+b}{2}}
\latex{ R }
\latex{ y }
\latex{ x }
Figure 81
\latex{y=x+\frac{1}{x}}
\latex{ -2 }
\latex{ -1 }
\latex{ 2 }
\latex{ 1 }
\latex{ y }
\latex{ x }
Let us select two arbitrary positive numbers, for example \latex{0\lt a\lt b} (Figure 80).
The \latex{y-}coordinate of the mid-point of the chord connecting the points P and Q of the curve
\latex{\frac{a+\dfrac{1}{a}+b+\dfrac{1}{b}}{2}=\frac{a+b}{2}+\frac{\dfrac{1}{a}+\dfrac{1}{b}}{2}}.
The \latex{y-}coordinate of the curve point R belonging to the mid-point of interval \latex{[a; b]}: \latex{\frac{a+b}{2}+\frac{1}{\dfrac{a+b}{2}}}.
The latter one is obviously less than the previous one. For this it is enough to show that \latex{\frac{\dfrac{1}{a}+\dfrac{1}{b}}{2}\gt \frac{1}{\dfrac{a+b}{2}}}, but we have shown it already when examining the convexity of the curve of, \latex{x\mapsto \frac{1}{x} }. The image of the whole function \latex{f} is shown in Figure 81.
It is worth observing that at negative x values the function has a maximum at \latex{ –1 }, and its maximum value is \latex{ –2 }.
Example 2
Let us plot the graph of and characterise function \latex{g: \R\rightarrow\R}, \latex{g(x)=\frac{1}{1+x^2}}.
Solution
It is again good to realise that \latex{ g } is an even function, because \latex{g(-x)=g(x)}, thus its curve is symmetric about the \latex{ y- }axis. So it is enough to examine when \latex{x\gt0}, and then reflect the resulting curve about the \latex{ y- }axis.
To draw the graph of function \latex{ g } more precisely we also need “finer”, higher-level mathematical tools.
\latex{y=x^2+1}
Figure 82
\latex{(x\geq0)}
\latex{y=\frac{1}{1+x^2}}
\latex{ 1 }
\latex{ x }
\latex{ y }
\latex{ 1 }
Now we choose the following way: we first plot the image of function \latex{x\mapsto x^2+1, \; x\geq 0}. It is the half of a normal parabola shifted up by \latex{ 1 }. Then we have to take its reciprocal. (Figure 82)
Since \latex{1\leq x^2+1, \; 0\lt \frac{1}{1+x^2}\leq1}, thus the curve will be between straight line \latex{y=1} and the \latex{x-}axis, and it will be decreasing, since \latex{x\mapsto x^2+1} was increasing. The image of the whole function is shown in Figure 83. The curve has its own name, it is called the witch of Agnesi.
\latex{y=\frac{1}{x^2+1}}
Figure 83
\latex{ -4 }
\latex{ -3 }
\latex{ -2 }
\latex{ -1 }
\latex{ 0 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ x }
\latex{ y }
Example 3
Let us plot the graph of and characterise function \latex{h:\R\rightarrow\R}, \latex{h(x)=\frac{2x}{1+x^2}}.
Solution
Function \latex{h} is odd because \latex{h(-x)=\frac{-2x}{1+(-x)^2}=-\frac{2x}{1+x^2}=-h(x)}.
So it is enough to examine for \latex{x\geq0}. \latex{h(0)=0}, after this we can assume that \latex{x \gt 0}. Let us divide both the numerator and the denominator of the fraction defining function \latex{h} by \latex{x}:
\latex{h(x)=\frac{2}{x+\dfrac{1}{x}}}.
It is easy to realise that this is exactly the reciprocal of \latex{\frac{x+\dfrac{1}{x}}{2}}.
Figure 84
\latex{y=\frac{2x}{1+x^2}}
\latex{(x\geq0)}
\latex{ 1 }
\latex{ 1 }
\latex{ x }
\latex{ y }
We know its double for positive values \latex{x}, so we can draw this curve too (Figure 84). The image of the whole function is a nice graph which has its own name: Newton's serpentine (Figure 85). To draw a more precise graph it is again necessary to use higher-level mathematical tools.
The function is decreasing on \latex{]–\infty; –1]}, it is increasing on \latex{[–1; 1]}, then it is decreasing on \latex{[1; +\infty[}. It has a minimum at \latex{ –1 }, its value is \latex{ –1 }, it has a maximum at \latex{ 1 }, its value is \latex{ 1 }. The range of the function is the closed interval \latex{[–1; 1]}.
\latex{y=\frac{2x}{1+x^2}}
Figure 85
\latex{ -1 }
\latex{ -1 }
\latex{ 1 }
\latex{ 1 }
\latex{ x }
\latex{ y }
Example 4
Let us plot the graph of and characterise function
\latex{k: (\R\backslash \left\{-1; 1\right\})\rightarrow \R}, \latex{k(x)=\frac{1}{1-x^2}}.
Figure 86
\latex{y=1-x^2}
\latex{(x\geq0)}
\latex{ 1 }
\latex{ 1 }
\latex{ x }
\latex{ y }
Solution
Function \latex{k} is an even function, so it is enough to plot the graph for \latex{x\geq0}, \latex{x\neq1}, then to reflect the resulting curve about the \latex{ y- }axis.
Let us first sketch the graph of function \latex{x\mapsto 1-x^2}, \latex{x\geq 0}, its image is half of a parabola. (Figure 86)
Let us first sketch the graph of function \latex{x\mapsto 1-x^2}, \latex{x\geq 0}, its image is half of a parabola. (Figure 86)
We can draw its reciprocal at places \latex{x\neq1}, \latex{x\gt 0}. (Figure 87)
Then we reflect the resulting curve about the \latex{ y- }axis. (Figure 88)
Then we reflect the resulting curve about the \latex{ y- }axis. (Figure 88)
The function \latex{ k } is decreasing on \latex{]–\infty; –1[} and on \latex{]–1; 0]}, it is increasing on \latex{[0; 1[} and on \latex{]1; +\infty[}. It has a local minimum at \latex{ 0 }, its minimum value is \latex{ 1 }.
Figure 87
\latex{y=\frac{1}{1-x^2}}
\latex{(x\geq0)}
\latex{ 1 }
\latex{ y }
\latex{ x }
\latex{ 1 }
Figure 88
\latex{y=\frac{1}{1-x^2}}
\latex{ -1 }
\latex{ 1 }
\latex{ 1 }
\latex{ y }
\latex{ x }
The range of \latex{k} is number set \latex{]-\infty; 0[ \cup[1; +\infty[}, it can also be given as follows: number set \latex{\R\backslash[0; 1[}.
Exercises
{{exercise_number}}. Plot the graphs of and characterise the following functions.
- \latex{f: (\R\backslash\left\{-1\right\})\rightarrow\R, \;\; f(x)=\frac{x^2}{x+1}}
- \latex{g: (\R\backslash\left\{1\right\})\rightarrow\R, \;\; g(x)=\frac{x^2-2x+2}{2x-2}}
- \latex{h: (\R\backslash\left\{0\right\})\rightarrow\R, \;\; h(x)=\frac{1}{x^2}}
- \latex{k: (\R\backslash\left\{2\right\})\rightarrow\R, \;\; k(x)=\frac{1}{x^2-4x+4}}
Quiz
The vases seen in the figure are empty at the beginning, and then we start filling them with water through one tap each. The water is flowing steadily from the taps (the same volume in a time unit).
Sketch the graph of the function which describes the change of the water surface area with time.
