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Let's study the function \(y= \frac(x^3)(1-x) \) and build its graph.
1. Scope of definition.
The domain of definition of a rational function (fraction) will be: the denominator is not equal to zero, i.e. \(1 -x \ne 0 => x \ne 1\). Domain $$D_f= (-\infty; 1) \cup (1;+\infty)$$
2. Function break points and their classification.
The function has one break point x = 1
Let's examine the point x= 1. Let's find the limit of the function to the right and left of the discontinuity point, to the right $$ \lim_(x \to 1+0) (\frac(x^3)(1-x)) = -\infty $$ and to the left of the point $$ \lim_(x \to 1-0)(\frac(x^3)(1-x)) = +\infty $$ This is a discontinuity point of the second kind because one-sided limits are equal to \(\infty\).
The straight line \(x = 1\) is a vertical asymptote.
3. Function parity.
We check for parity \(f(-x) = \frac((-x)^3)(1+x) \) the function is neither even nor odd.
4. Zeros of the function (points of intersection with the Ox axis). Intervals of constant sign of a function.
Function zeros ( point of intersection with the Ox axis): we equate \(y=0\), we get \(\frac(x^3)(1-x) = 0 => x=0 \). The curve has one intersection point with the Ox axis with coordinates \((0;0)\).
Intervals of constant sign of a function.
On the considered intervals \((-\infty; 1) \cup (1;+\infty)\) the curve has one point of intersection with the Ox axis, so we will consider the domain of definition on three intervals.
Let us determine the sign of the function on intervals of the domain of definition:
interval \((-\infty; 0) \) find the value of the function at any point \(f(-4) = \frac(x^3)(1-x)< 0 \), на этом интервале функция отрицательная \(f(x) < 0 \), т.е. находится ниже оси Ox
interval \((0; 1) \) we find the value of the function at any point \(f(0.5) = \frac(x^3)(1-x) > 0 \), on this interval the function is positive \(f(x ) > 0 \), i.e. is located above the Ox axis.
interval \((1;+\infty) \) find the value of the function at any point \(f(4) = \frac(x^3)(1-x)< 0 \), на этом интервале функция отрицательная \(f(x) < 0 \), т.е. находится ниже оси Ox
5. Intersection points with the Oy axis: we equate \(x=0\), we get \(f(0) = \frac(x^3)(1-x) = 0\). Coordinates of the point of intersection with the Oy axis \((0; 0)\)
6. Intervals of monotony. Extrema of a function.
Let's find the critical (stationary) points, for this we find the first derivative and equate it to zero $$ y" = (\frac(x^3)(1-x))" = \frac(3x^2(1-x) + x^3)( (1-x)^2) = \frac(x^2(3-2x))( (1-x)^2) $$ equal to 0 $$ \frac(x^2(3 -2x))( (1-x)^2) = 0 => x_1 = 0 \quad x_2= \frac(3)(2)$$ Let's find the value of the function at this point \(f(0) = 0\) and \(f(\frac(3)(2)) = -6.75\). We got two critical points with coordinates \((0;0)\) and \((1.5;-6.75)\)
Intervals of monotony.
The function has two critical points (possible extremum points), so we will consider monotonicity on four intervals:
interval \((-\infty; 0) \) find the value of the first derivative at any point in the interval \(f(-4) = \frac(x^2(3-2x))( (1-x)^2) >
interval \((0;1)\) we find the value of the first derivative at any point in the interval \(f(0.5) = \frac(x^2(3-2x))( (1-x)^2) > 0\) , the function increases over this interval.
interval \((1;1.5)\) we find the value of the first derivative at any point in the interval \(f(1.2) = \frac(x^2(3-2x))( (1-x)^2) > 0\) , the function increases over this interval.
interval \((1.5; +\infty)\) find the value of the first derivative at any point in the interval \(f(4) = \frac(x^2(3-2x))( (1-x)^2)< 0\), на этом интервале функция убывает.
Extrema of a function.
When studying the function, we obtained two critical (stationary) points on the interval of the domain of definition. Let's determine whether they are extremes. Let us consider the change in the sign of the derivative when passing through critical points:
point \(x = 0\) the derivative changes sign with \(\quad +\quad 0 \quad + \quad\) - the point is not an extremum.
point \(x = 1.5\) the derivative changes sign with \(\quad +\quad 0 \quad - \quad\) - the point is a maximum point.
7. Intervals of convexity and concavity. Inflection points.
To find the intervals of convexity and concavity, we find the second derivative of the function and equate it to zero $$y"" = (\frac(x^2(3-2x))( (1-x)^2))"= \frac(2x (x^2-3x+3))((1-x)^3) $$Equate to zero $$ \frac(2x(x^2-3x+3))((1-x)^3)= 0 => 2x(x^2-3x+3) =0 => x=0$$ The function has one critical point of the second kind with coordinates \((0;0)\).
Let us define convexity on intervals of the domain of definition, taking into account a critical point of the second kind (a point of possible inflection).
interval \((-\infty; 0)\) find the value of the second derivative at any point \(f""(-4) = \frac(2x(x^2-3x+3))((1-x)^ 3)< 0 \), на этом интервале вторая производная функции отрицательная \(f""(x) < 0 \) - функция выпуклая вверх (вогнутая).
interval \((0; 1)\) we find the value of the second derivative at any point \(f""(0.5) = \frac(2x(x^2-3x+3))((1-x)^3) > 0 \), on this interval the second derivative of the function is positive \(f""(x) > 0 \) the function is convex downward (convex).
interval \((1; \infty)\) find the value of the second derivative at any point \(f""(4) = \frac(2x(x^2-3x+3))((1-x)^3)< 0 \), на этом интервале вторая производная функции отрицательная \(f""(x) < 0 \) - функция выпуклая вверх (вогнутая).
Inflection points.
Let us consider the change in the sign of the second derivative when passing through a critical point of the second kind:
At the point \(x =0\), the second derivative changes sign with \(\quad - \quad 0 \quad + \quad\), the graph of the function changes convexity, i.e. this is the inflection point with coordinates \((0;0)\).
8. Asymptotes.
Vertical asymptote. The graph of the function has one vertical asymptote \(x =1\) (see paragraph 2).
Oblique asymptote.
In order for the graph of the function \(y= \frac(x^3)(1-x) \) at \(x \to \infty\) to have a slanted asymptote \(y = kx+b\), it is necessary and sufficient , so that there are two limits $$\lim_(x \to +\infty)=\frac(f(x))(x) =k $$we find it $$ \lim_(x \to \infty) (\frac( x^3)(x(1-x))) = \infty => k= \infty $$ and the second limit $$ \lim_(x \to +\infty)(f(x) - kx) = b$ $, because \(k = \infty\) - there is no oblique asymptote.
Horizontal asymptote: in order for a horizontal asymptote to exist, it is necessary that there be a limit $$\lim_(x \to \infty)f(x) = b$$ let's find it $$ \lim_(x \to +\infty)(\frac( x^3)(1-x))= -\infty$$$$ \lim_(x \to -\infty)(\frac(x^3)(1-x))= -\infty$$
There is no horizontal asymptote.
9. Function graph.
The study of a function is carried out according to a clear scheme and requires the student to have a solid knowledge of basic mathematical concepts such as the domain of definition and values, continuity of the function, asymptote, extremum points, parity, periodicity, etc. The student must be able to differentiate functions freely and solve equations, which can sometimes be very complex.
That is, this task tests a significant layer of knowledge, any gap in which will become an obstacle to obtaining the correct solution. Particularly often, difficulties arise with constructing graphs of functions. This mistake is immediately noticeable to the teacher and can greatly damage your grade, even if everything else was done correctly. Here you can find online function research problems: study examples, download solutions, order assignments.
Explore a function and plot a graph: examples and solutions online
We have prepared for you a lot of ready-made function studies, both paid in the solution book and free in the section Examples of function studies. Based on these solved tasks, you will be able to familiarize yourself in detail with the methodology for performing similar tasks, and carry out your research by analogy.
We offer ready-made examples of complete research and plotting of functions of the most common types: polynomials, fractional-rational, irrational, exponential, logarithmic, trigonometric functions. Each solved problem is accompanied by a ready-made graph with highlighted key points, asymptotes, maxima and minima; the solution is carried out using an algorithm for studying the function.
In any case, the solved examples will be of great help to you as they cover the most popular types of functions. We offer you hundreds of already solved problems, but, as you know, there are an infinite number of mathematical functions in the world, and teachers are great experts at inventing more and more tricky tasks for poor students. So, dear students, qualified help will not hurt you.
Solving custom function research problems
In this case, our partners will offer you another service - full function research online to order. The task will be completed for you in compliance with all the requirements for an algorithm for solving such problems, which will greatly please your teacher.
We will do a complete study of the function for you: we will find the domain of definition and the domain of values, examine for continuity and discontinuity, establish parity, check your function for periodicity, and find the points of intersection with the coordinate axes. And, of course, further using differential calculus: we will find asymptotes, calculate extrema, inflection points, and construct the graph itself.
If the problem requires a complete study of the function f (x) = x 2 4 x 2 - 1 with the construction of its graph, then we will consider this principle in detail.
To solve a problem of this type, you should use the properties and graphs of basic elementary functions. The research algorithm includes the following steps:
Yandex.RTB R-A-339285-1
Finding the domain of definition
Since research is carried out on the domain of definition of the function, it is necessary to start with this step.
Example 1
The given example involves finding the zeros of the denominator in order to exclude them from the ODZ.
4 x 2 - 1 = 0 x = ± 1 2 ⇒ x ∈ - ∞ ; - 1 2 ∪ - 1 2 ; 1 2 ∪ 1 2 ; +∞
As a result, you can get roots, logarithms, and so on. Then the ODZ can be searched for a root of an even degree of type g (x) 4 by the inequality g (x) ≥ 0, for the logarithm log a g (x) by the inequality g (x) > 0.
Studying the boundaries of the ODZ and finding vertical asymptotes
There are vertical asymptotes at the boundaries of the function, when the one-sided limits at such points are infinite.
Example 2
For example, consider the border points equal to x = ± 1 2.
Then it is necessary to study the function to find the one-sided limit. Then we get that: lim x → - 1 2 - 0 f (x) = lim x → - 1 2 - 0 x 2 4 x 2 - 1 = = lim x → - 1 2 - 0 x 2 (2 x - 1 ) (2 x + 1) = 1 4 (- 2) · - 0 = + ∞ lim x → - 1 2 + 0 f (x) = lim x → - 1 2 + 0 x 2 4 x - 1 = = lim x → - 1 2 + 0 x 2 (2 x - 1) (2 x + 1) = 1 4 (- 2) (+ 0) = - ∞ lim x → 1 2 - 0 f (x) = lim x → 1 2 - 0 x 2 4 x 2 - 1 = = lim x → 1 2 - 0 x 2 (2 x - 1) (2 x + 1) = 1 4 (- 0) 2 = - ∞ lim x → 1 2 - 0 f (x) = lim x → 1 2 - 0 x 2 4 x 2 - 1 = = lim x → 1 2 - 0 x 2 (2 x - 1) (2 x + 1) = 1 4 ( + 0) 2 = + ∞
This shows that the one-sided limits are infinite, which means the straight lines x = ± 1 2 are the vertical asymptotes of the graph.
Study of a function and whether it is even or odd
When the condition y (- x) = y (x) is satisfied, the function is considered even. This suggests that the graph is located symmetrically with respect to Oy. When the condition y (- x) = - y (x) is satisfied, the function is considered odd. This means that the symmetry is relative to the origin of coordinates. If at least one inequality is not satisfied, we obtain a function of general form.
The equality y (- x) = y (x) indicates that the function is even. When constructing, it is necessary to take into account that there will be symmetry with respect to Oy.
To solve the inequality, intervals of increasing and decreasing are used with the conditions f " (x) ≥ 0 and f " (x) ≤ 0, respectively.
Definition 1
Stationary points- these are the points that turn the derivative to zero.
Critical points- these are internal points from the domain of definition where the derivative of the function is equal to zero or does not exist.
When making a decision, the following notes must be taken into account:
- for existing intervals of increasing and decreasing inequalities of the form f " (x) > 0, critical points are not included in the solution;
- points at which the function is defined without a finite derivative must be included in the intervals of increasing and decreasing (for example, y = x 3, where the point x = 0 makes the function defined, the derivative has the value of infinity at this point, y " = 1 3 x 2 3, y "(0) = 1 0 = ∞, x = 0 is included in the increasing interval);
- To avoid disagreements, it is recommended to use mathematical literature recommended by the Ministry of Education.
Inclusion of critical points in intervals of increasing and decreasing if they satisfy the domain of definition of the function.
Definition 2
For determining the intervals of increase and decrease of a function, it is necessary to find:
- derivative;
- critical points;
- divide the definition domain into intervals using critical points;
- determine the sign of the derivative on each of the intervals, where + is an increase and - is a decrease.
Example 3
Find the derivative on the domain of definition f " (x) = x 2 " (4 x 2 - 1) - x 2 4 x 2 - 1 " (4 x 2 - 1) 2 = - 2 x (4 x 2 - 1) 2 .
Solution
To solve you need:
- find stationary points, this example has x = 0;
- find the zeros of the denominator, the example takes the value zero at x = ± 1 2.
We place points on the number line to determine the derivative on each interval. To do this, it is enough to take any point from the interval and perform a calculation. If the result is positive, we depict + on the graph, which means the function is increasing, and - means it is decreasing.
For example, f " (- 1) = - 2 · (- 1) 4 - 1 2 - 1 2 = 2 9 > 0, which means that the first interval on the left has a + sign. Consider on the number line.
Answer:
- the function increases on the interval - ∞; - 1 2 and (- 1 2 ; 0 ] ;
- there is a decrease in the interval [ 0 ; 1 2) and 1 2 ; + ∞ .
In the diagram, using + and -, the positivity and negativity of the function are depicted, and the arrows indicate decrease and increase.
Extremum points of a function are points where the function is defined and through which the derivative changes sign.
Example 4
If we consider an example where x = 0, then the value of the function in it is equal to f (0) = 0 2 4 · 0 2 - 1 = 0. When the sign of the derivative changes from + to - and passes through the point x = 0, then the point with coordinates (0; 0) is considered the maximum point. When the sign changes from - to +, we obtain a minimum point.
Convexity and concavity are determined by solving inequalities of the form f "" (x) ≥ 0 and f "" (x) ≤ 0. Less commonly used is the name convexity down instead of concavity, and convexity upward instead of convexity.
Definition 3
For determining the intervals of concavity and convexity necessary:
- find the second derivative;
- find the zeros of the second derivative function;
- divide the definition area into intervals with the appearing points;
- determine the sign of the interval.
Example 5
Find the second derivative from the domain of definition.
Solution
f "" (x) = - 2 x (4 x 2 - 1) 2 " = = (- 2 x) " (4 x 2 - 1) 2 - - 2 x 4 x 2 - 1 2 " (4 x 2 - 1) 4 = 24 x 2 + 2 (4 x 2 - 1) 3
We find the zeros of the numerator and denominator, where in our example we have that the zeros of the denominator x = ± 1 2
Now you need to plot the points on the number line and determine the sign of the second derivative from each interval. We get that
Answer:
- the function is convex from the interval - 1 2 ; 12 ;
- the function is concave from the intervals - ∞ ; - 1 2 and 1 2; + ∞ .
Definition 4
Inflection point– this is a point of the form x 0 ; f (x 0) . When it has a tangent to the graph of the function, then when it passes through x 0 the function changes sign to the opposite.
In other words, this is a point through which the second derivative passes and changes sign, and at the points themselves it is equal to zero or does not exist. All points are considered to be the domain of the function.
In the example, it was clear that there are no inflection points, since the second derivative changes sign while passing through the points x = ± 1 2. They, in turn, are not included in the scope of definition.
Finding horizontal and oblique asymptotes
When defining a function at infinity, you need to look for horizontal and oblique asymptotes.
Definition 5
Oblique asymptotes are depicted using straight lines given by the equation y = k x + b, where k = lim x → ∞ f (x) x and b = lim x → ∞ f (x) - k x.
For k = 0 and b not equal to infinity, we find that the oblique asymptote becomes horizontal.
In other words, asymptotes are considered to be lines to which the graph of a function approaches at infinity. This facilitates quick construction of a function graph.
If there are no asymptotes, but the function is defined at both infinities, it is necessary to calculate the limit of the function at these infinities in order to understand how the graph of the function will behave.
Example 6
Let's consider as an example that
k = lim x → ∞ f (x) x = lim x → ∞ x 2 4 x 2 - 1 x = 0 b = lim x → ∞ (f (x) - k x) = lim x → ∞ x 2 4 x 2 - 1 = 1 4 ⇒ y = 1 4
is a horizontal asymptote. After examining the function, you can begin to construct it.
Calculating the value of a function at intermediate points
To make the graph more accurate, it is recommended to find several function values at intermediate points.
Example 7
From the example we considered, it is necessary to find the values of the function at the points x = - 2, x = - 1, x = - 3 4, x = - 1 4. Since the function is even, we get that the values coincide with the values at these points, that is, we get x = 2, x = 1, x = 3 4, x = 1 4.
Let's write and solve:
F (- 2) = f (2) = 2 2 4 2 2 - 1 = 4 15 ≈ 0, 27 f (- 1) - f (1) = 1 2 4 1 2 - 1 = 1 3 ≈ 0 , 33 f - 3 4 = f 3 4 = 3 4 2 4 3 4 2 - 1 = 9 20 = 0 , 45 f - 1 4 = f 1 4 = 1 4 2 4 1 4 2 - 1 = - 1 12 ≈ - 0.08
To determine the maxima and minima of the function, inflection points, and intermediate points, it is necessary to construct asymptotes. For convenient designation, intervals of increasing, decreasing, convexity, and concavity are recorded. Let's look at the picture below.
It is necessary to draw graph lines through the marked points, which will allow you to approach the asymptotes by following the arrows.
This concludes the full exploration of the function. There are cases of constructing some elementary functions for which geometric transformations are used.
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Conduct a complete study and graph the function
y(x)=x2+81−x.y(x)=x2+81−x.
1) The scope of the function. Since the function is a fraction, we need to find the zeros of the denominator.
1−x=0,⇒x=1.1−x=0,⇒x=1.
We exclude the only point x=1x=1 from the domain of definition of the function and get:
D(y)=(−∞;1)∪(1;+∞).D(y)=(−∞;1)∪(1;+∞).
2) Let us study the behavior of the function in the vicinity of the discontinuity point. Let's find one-sided limits:
Since the limits are equal to infinity, the point x=1x=1 is a discontinuity of the second kind, the straight line x=1x=1 is a vertical asymptote.
3) Let us determine the intersection points of the function graph with the coordinate axes.
Let's find the points of intersection with the ordinate axis OyOy, for which we equate x=0x=0:
Thus, the point of intersection with the OyOy axis has coordinates (0;8)(0;8).
Let's find the points of intersection with the abscissa axis OxOx, for which we set y=0y=0:
The equation has no roots, so there are no points of intersection with the OxOx axis.
Note that x2+8>0x2+8>0 for any xx. Therefore, for x∈(−∞;1)x∈(−∞;1), the function y>0y>0 (takes positive values, the graph is above the x-axis), for x∈(1;+∞)x∈(1; +∞) function y<0y<0 (принимает отрицательные значения, график находится ниже оси абсцисс).
4) The function is neither even nor odd because:
5) Let's examine the function for periodicity. The function is not periodic, since it is a fractional rational function.
6) Let's examine the function for extrema and monotonicity. To do this, we find the first derivative of the function:
Let's equate the first derivative to zero and find stationary points (at which y′=0y′=0):
We got three critical points: x=−2,x=1,x=4x=−2,x=1,x=4. Let us divide the entire domain of definition of the function into intervals with these points and determine the signs of the derivative in each interval:
For x∈(−∞;−2),(4;+∞)x∈(−∞;−2),(4;+∞) the derivative y′<0y′<0, поэтому функция убывает на данных промежутках.
For x∈(−2;1),(1;4)x∈(−2;1),(1;4) the derivative y′>0y′>0, the function increases on these intervals.
In this case, x=−2x=−2 is a local minimum point (the function decreases and then increases), x=4x=4 is a local maximum point (the function increases and then decreases).
Let's find the values of the function at these points: 
Thus, the minimum point is (−2;4)(−2;4), the maximum point is (4;−8)(4;−8).
7) Let's examine the function for kinks and convexity. Let's find the second derivative of the function:
Let us equate the second derivative to zero:
The resulting equation has no roots, so there are no inflection points. Moreover, when x∈(−∞;1)x∈(−∞;1) y′′>0y″>0 is satisfied, that is, the function is concave, when x∈(1;+∞)x∈(1;+ ∞) is satisfied by y′′<0y″<0, то есть функция выпуклая.
8) Let us examine the behavior of the function at infinity, that is, at .
Since the limits are infinite, there are no horizontal asymptotes.
Let's try to determine oblique asymptotes of the form y=kx+by=kx+b. We calculate the values of k,bk,b using known formulas:
We found that the function has one oblique asymptote y=−x−1y=−x−1.
9) Additional points. Let's calculate the value of the function at some other points in order to more accurately construct the graph.
y(−5)=5.5;y(2)=−12;y(7)=−9.5.y(−5)=5.5;y(2)=−12;y(7)=−9.5.
10) Based on the data obtained, we will construct a graph, supplement it with asymptotes x=1x=1 (blue), y=−x−1y=−x−1 (green) and mark the characteristic points (purple intersection with the ordinate axis, orange extrema, black additional points) :
Task 4: Geometric, Economic problems (I have no idea what, here is an approximate selection of problems with solutions and formulas) 
Example 3.23. a
Solution. x And y y
y = a - 2×a/4 =a/2. Since x = a/4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. For xa/4 S " > 0, and for x >a/4 S "< 0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед).Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.
Example 3.24.
Solution.
R = 2, H = 16/4 = 4.
Example 3.22. Find the extrema of the function f(x) = 2x 3 - 15x 2 + 36x - 14.
Solution. Since f "(x) = 6x 2 - 30x +36 = 6(x -2)(x - 3), then the critical points of the function x 1 = 2 and x 2 = 3. Extrema can only be at these points. So as when passing through the point x 1 = 2 the derivative changes its sign from plus to minus, then at this point the function has a maximum. When passing through the point x 2 = 3 the derivative changes its sign from minus to plus, therefore at the point x 2 = 3 the function has a minimum. Having calculated the function values at the points
x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f(2) = 14 and minimum f(3) = 13.
Example 3.23. It is necessary to build a rectangular area near the stone wall so that it is fenced off on three sides with wire mesh, and the fourth side is adjacent to the wall. For this there is a linear meters of mesh. At what aspect ratio will the site have the largest area?
Solution. Let us denote the sides of the platform by x And y. The area of the site is S = xy. Let y- this is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must hold. Therefore y = a - 2x and S = x(a - 2x), where
0 ≤ x ≤ a/2 (the length and width of the pad cannot be negative). S " = a - 4x, a - 4x = 0 at x = a/4, whence
y = a - 2×a/4 =a/2. Since x = a/4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. For xa/4 S " > 0, and for x >a/4 S "< 0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед).Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.
Example 3.24. It is required to manufacture a closed cylindrical tank with a capacity V=16p ≈ 50 m 3 . What should be the dimensions of the tank (radius R and height H) so that the least amount of material is used for its manufacture?
Solution. The total surface area of the cylinder is S = 2pR(R+H). We know the volume of the cylinder V = pR 2 N Þ N = V/pR 2 =16p/ pR 2 = 16/ R 2 . This means S(R) = 2p(R 2 +16/R). We find the derivative of this function:
S " (R) = 2p(2R- 16/R 2) = 4p (R- 8/R 2). S " (R) = 0 for R 3 = 8, therefore,
R = 2, H = 16/4 = 4.
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