Instructions for mathematical analysis

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Given the function f left parenthesis x right parenthesis equals fraction numerator 2 x cubed minus 5 x squared plus 4 x plus 1 over denominator 2 x squared minus x minus 1 end fraction, here's an exercise to find a series of properties that will allow you to represent it graphically by hand. Academically, it will help you review some basic commands related to the world of analysis.

To solve the first question, you must use the command domain and the icon you can find in the Calculus section of the Menu to calculate the limits. Notice how the points outside the domain of the function coincide with the roots of the denominator.

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The second question is easier to solve with the tools seen in the previous question. The fact this limit equals 0 guarantees the line y equals x minus 2 is an oblique asymptote of the initial function.

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To solve the third question, you simply have to derive the initial function using the apostrophe. If you prefer it, you can also find it by using the action that you can find in the top bar or through the differentiate command. Note that if you want to calculate these derivatives by hand you must use the derivation rule of a quotient.

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You can solve the equations by equating the expression to 0 or using the solve command interchangeably. Naming the solutions allows you to save them in memory and use them in subsequent calculations using more decimal places than CalcMe shows on the screen.

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The fifth question is solved by evaluating the previously found second derivative of the function at the critical points. If this is positive on the point, it's a relative minimum. On the contrary, if it's negative, it will be a relative maximum.

On the other hand, if you want to find the intervals of increase and decrease, you should see when the derived function is negative (decreasing function) and when it's positive (increasing function). These intervals are delimited by relative extremes and points outside the domain.

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As for the inflection points, just make sure the point nullifying the second derivative does not nullify the third one as well. Therefore, c equals 0.27788 will be an inflection point.

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Finally, you need to represent the function and its asymptotes.

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Given a set of functions, you can use CalcMe to find and draw the area of the region between them. To do this you will need to find the integration limits and see which function is greater in the related range.

Given the functions y equals sin left parenthesis x right parenthesis and y equals 2 times sin left parenthesis x right parenthesis times cos left parenthesis x right parenthesis, you want to find the area of the region between them from the origin to the first point where they intersect with a positive abscissa. When looking for the cut-off points, you can see the first is x equals straight pi over 3, therefore you will have to calculate the integral between 0 and straight pi over 3 of the function that goes above (which is y equals 2 times sin left parenthesis x right parenthesis times cos left parenthesis x right parenthesis) minus the one that goes below (y equals sin left parenthesis x right parenthesis).

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Given this time the functions y equals negative x squared plus 2 x and y equals x squared minus 2 x plus 2, you want to find the region area between the OY axis (x equals 0) and both functions. As before, you have to find its intersection and calculate the integral between 0 and this point. This time, y equals x squared minus 2 x plus 2 goes over and y equals negative x squared plus 2 x, above.

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If you want to represent these regions, you have two choices. If the region you want to plot is encapsulated between two functions, you just have to use the region command and indicate the functions and the range you want to represent.

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On the other hand, if the represented region is not properly between two functions, you can directly intersect the necessary inequalities and draw it using the action.

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Given any function, with a wide variety of purposes, you may be interested in finding the maximum and minimum in a given interval. With CalcMe you can easily find them; you simply have to evaluate the function at its critical points and at the extremes of the range of possible values.

Given a food price index model I left parenthesis t right parenthesis equals 0.00009045 times t to the power of 5plus 0.001438 times t to the power of 4negative 0.0656 times t cubedplus 0.4598 times t squarednegative 0.6270 times tplus 99.33, you want to find when in the first six years the food has been cheaper and more expensive. When you look for the derivative and equating it to zero, you will see there are points outside our range that you will not have to take into account. Therefore, it remains only to evaluate the function at the points in question and see when it is bigger and when it is smaller.

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Repeating the same exercise with a period of 12 years, you have to add x equals 11.043 to the list of optimal candidates. This way you can see how the minimum is achieved later than in the first case.

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Given any function, you can use CalcMe to find its Taylor polynomial of a specific degree centered on a given point. To do this, you can use the formula or the command taylor.

Given the function g left parenthesis x right parenthesis equals ln left parenthesis 2 minus x squared right parenthesis, let's first calculate the Taylor polynomial of degree 2 centered at x equals 1. You can use the formula and CalcMe will directly find the coefficients.

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Given the functions g left parenthesis x right parenthesis equals ln left parenthesis 2 minus x squared right parenthesis, now let's calculate the Taylor polynomial of degree 7 centered on the same point. Since the formula is very long, you can choose to use the taylor command indicating the function, the variable, the point, and the degree. INTERACTIVE DEMO

After looking at the two approaches, you can consider representing them graphically to see how they relate to the initial function.

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Note the higher the degree of the Taylor polynomial, the greater the interval at which the function approximates properly. In this example, the blue function adequately approximates the polynomial to the interval left parenthesis 0.8 comma 1.2 right parenthesis while the red one approaches the interval left parenthesis 0.6 comma 1.4 right parenthesis.

As you have seen before, given a set of functions you can use the calculator to find the area of the region you are delineating. The usual process is to use the Barrow rule. That is, find a primitive of the function in question and evaluate it at the extremes of integration.

Sometimes, when these integration limits approach a non-real number (plus infinity and negative infinity), we talk about an improper integral and calculate the limit of a definite integral.

Firstly, given the curve y equals left parenthesis square root of 5 minus square root of x right parenthesis squared, calculate the cut-off point and the integral between 0 and the point of our function to find the area of the related region.

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Secondly, given the curve y equals fraction numerator x squared minus 1 over denominator x squared plus 1 end fraction, calculate the integral with extremes of integration minus and plus infinity of the horizontal asymptote minus the function to find the area of the related function. Since integration bounds approach a specific non-real number (plus infinity and negative infinity), we are talking about an improper integral.

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Finally, you can calculate the area of the bounded region between the two curves above.

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Lastly, CalcMe also lets you solve ODEs with or without initial conditions using the solve command. When writing the equation, you should consider entering the function using the icon in the Calculus section of the Menu. Otherwise, CalcMe will not understand you are talking about a differential equation.

Thus, once the initial conditions are given, CalcMe returns the equation of a variable that satisfies the ODE and the initial conditions in question.

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