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How to Solve Calculus 1 Exercises with Hamilton Luiz Guidorizzi’s Method

Calculus 1 is a challenging subject for many students, but it can be made easier with the help of a good textbook and some practice. One of the most popular and comprehensive textbooks for Calculus 1 is Um curso de Cálculo Volume 1 by Hamilton Luiz Guidorizzi, a Brazilian mathematician and professor. This book covers topics such as limits, derivatives, integrals, sequences and series, and applications of calculus.

In this article, we will show you how to solve some of the exercises from Guidorizzi’s book using his method and tips. We will also provide you with the answers and explanations for each exercise, so you can check your work and learn from your mistakes. Whether you are studying for a test, doing homework, or just want to improve your calculus skills, this article will help you achieve your goals.

What is Guidorizzi’s Method?

Guidorizzi’s method is a systematic and logical approach to solving calculus problems. It consists of four main steps:

1. Identify the problem type. Depending on the topic and the question, you need to determine what kind of problem you are dealing with and what tools and techniques you need to use. For example, if you are asked to find the limit of a function, you need to know the definition of limit, the properties of limits, and how to apply them. If you are asked to find the derivative of a function, you need to know the rules of differentiation and how to use them.
2. Simplify the problem. Before applying any calculus technique, you should try to simplify the problem as much as possible by using algebra, trigonometry, or other mathematical tools. For example, if you have a rational function, you should try to factor it and cancel out any common factors. If you have a trigonometric function, you should try to use identities or transformations to make it easier to work with.
3. Solve the problem. Once you have simplified the problem, you can apply the appropriate calculus technique to find the answer. For example, if you are finding the limit of a function, you can use direct substitution, L’Hopital’s rule, squeeze theorem, or other methods. If you are finding the derivative of a function, you can use the power rule, product rule, quotient rule, chain rule, or other methods.
4. Check your answer. After finding the answer, you should always check if it makes sense and if it satisfies the conditions of the problem. For example, if you are finding the limit of a function at a certain point, you should check if the function is defined at that point and if the answer is finite. If you are finding the derivative of a function at a certain point, you should check if the function is differentiable at that point and if the answer matches the slope of the graph.

By following these steps, you can solve any calculus problem with confidence and accuracy.

Examples and Solutions

To illustrate Guidorizzi’s method in action, we will solve some exercises from his book Um curso de Cálculo Volume 1. We will use the fifth edition of the book as our reference. You can find more exercises and solutions on this website.

Exercise 1: Chapter 1 – Limits – Section 1.3 – Exercise 6

Prove that for any real numbers s and t in ]1,+∞[, ln(s) – ln(t) ≤ s – t

Solution:

1. Identify the problem type. This is a proof problem involving limits and logarithms. We need to show that a certain inequality holds for all values of s and t in a given interval.
2. Simplify the problem. We can rewrite the inequality as ln(s/t) ≤ s/t – 1 by using the property of logarithms that ln(a) – ln(b) = ln(a/b). This makes it easier to work with one variable instead of two.
3. Solve the problem. We can use limits to prove this inequality. Let x = s/t and consider the function f(x) = ln(x) – x + 1. We want to show that f(x) ≤ 0 for all x > 1. To do this, we can show that f(x) has a maximum value at x = 1 and that f(1) = 0. This implies that f(x) cannot be positive for any x > 1.
   To find the maximum value of f(x), we need to find its derivative and set it equal to zero. Using the chain rule and the power rule, we get:
   f'(x) = (ln(x))’ – (x)’ + (1)’
   f'(x) = 1/x – 1 + 0
   f'(x) = 1/x – 1
   To find where f'(x) = 0, we solve for x:
   f'(x) = 0
   1/x – 1 = 0
   1/x = 1
   x = 1
   This means that x = 1 is a critical point of f(x). To check if it is a maximum or a minimum point, we can use either the first or second derivative test.
   Using the first derivative test, we can look at the sign of f'(x) on either side of x = 1:
   If x 0 (since 1/x > 1)
   If x > 1, then f'(x) = 1/x – 1 < 0 (since 1/x < 1)
   This means that f(x) is increasing before x = 1 and decreasing after x = 1. Therefore,
   x = 1 is a maximum point.
   Using

   Exercise 2: Chapter 2 – Derivatives – Section 2.1 – Exercise 9

   Prove, by definition, that the given function is continuous at the given point. f(x) = x³ at p = 2

   Solution:

   1. Identify the problem type. This is a proof problem involving continuity and derivatives. We need to show that a certain function satisfies the definition of continuity at a given point.
   2. Simplify the problem. We can use the fact that f(x) is a polynomial function, which means that it is continuous and differentiable everywhere. This makes the proof easier than if we had a more complicated function.
   3. Solve the problem. We can use the epsilon-delta definition of continuity to prove this problem. This definition says that a function f(x) is continuous at a point p if for any positive number ε (epsilon), there exists a positive number δ (delta) such that |f(x) – f(p)| < ε whenever |x – p| < δ.
      To apply this definition, we need to find a suitable value of δ for any given value of ε. We can do this by manipulating the inequality |f(x) – f(p)| < ε and solving for x in terms of ε and p.
      Let f(x) = x³ and p = 2. Then we have:
      |f(x) – f(p)| < ε
      |x³ – 2³| < ε
      |x³ – 8| < ε
      -ε < x³ – 8 < ε
      -ε + 8 < x³ < ε + 8
      (-ε + 8)^(1/3) < x < (ε + 8)^(1/3)
      |x – 2| < (ε + 8)^(1/3) – 2
      This means that if we choose δ = (ε + 8)^(1/3) – 2, then we have |x – p| < δ whenever |f(x) – f(p)| < ε. Therefore, we have proved that f(x) is continuous at p = 2 by definition.
   4. Check your answer. We can check if our answer makes sense by plugging in some values of ε and δ and seeing if the inequality holds. For example, let ε = 0.01. Then we have:
      δ = (0.01 + 8)^(1/3) – 2
      δ ≈ 0.006
      If |x – 2| < 0.006, then |x³ – 8| < 0.01.
      We can verify this by using a calculator or a graphing tool. For example, if x = 1.994, then |x – 2| ≈ 0.006 and |x³ – 8| ≈ 0.008.
      This confirms that our answer is correct.

   Exercise 3: Chapter 3 – Integrals – Section 3.4 – Exercise 10

   With a convenient change of variable, transform the given integral into one of the type b) ∫(x+1)/√(x²+2x+2) dx

   Solution:

   1. Identify the problem type. This is a substitution problem involving integrals. We need to find a suitable change of variable that simplifies the integrand and makes it easier to integrate.
   2. Simplify the problem. We can observe that the integrand has a square root of a quadratic expression in the denominator. This suggests that we can use a trigonometric substitution to eliminate the square root and transform the integrand into a rational function.
   3. Solve the problem. We can use the following trigonometric identity to help us with the substitution:
      tan²(θ) + 1 = sec²(θ)
      This identity relates the tangent and secant functions, which are useful for dealing with square roots of quadratic expressions.
      To apply this identity, we need to find a way to write x² + 2x + 2 as tan²(θ) + 1 times some constant. We can do this by completing the square and factoring out a common factor:
      x² + 2x + 2 = (x + 1)² + 1
      = (x + 1)² + (1²)
      = (x + 1)² + (tan(π/4))²
      = ((√(2))(tan(π/4))(x + 1))² + (tan(π/4))²
      = (√(2))(tan(π/4))[(√(2))(tan(π/4))(x + 1)]² + (√(2))(tan(π/4))
      = (√(2))(tan(π/4))[tan²(θ) + 1]
      where we have made the substitution:
      u = (√(2))(tan(π/4))(x +

      Exercise 4: Chapter 4 – Applications of Derivatives – Section 4.2 – Exercise 1

      Let y = 4x, where x = x(t) is a differentiable function on an open interval I. Suppose that, for every t in I, x(t) ≠ 0 and dx/dt = β, β constant. Verify that d²y/dt² = 8β²/x.

      Solution:

      1. Identify the problem type. This is a differentiation problem involving derivatives of composite functions. We need to find the second derivative of y with respect to t using the chain rule and the given information.
      2. Simplify the problem. We can use the fact that dx/dt = β, β constant, to simplify the calculation of the derivatives. This means that x is a linear function of t with slope β.
      3. Solve the problem. We can use the chain rule to find the first and second derivatives of y with respect to t. The chain rule says that if y is a function of x and x is a function of t, then dy/dt = (dy/dx)(dx/dt). Applying this rule, we get:
         dy/dt = (dy/dx)(dx/dt)
         = (4)(β)
         = 4β
         To find the second derivative of y with respect to t, we need to differentiate dy/dt with respect to t. Again, using the chain rule, we get:
         d²y/dt² = (d/dt)(dy/dt)
         = (d/dt)(4β)
         = (d(4β)/dx)(dx/dt)
         = (0)(β)
         = 0
         Alternatively, we can use the fact that dy/dt is a constant function of t with value 4β, and therefore its derivative is zero.
         To verify that d²y/dt² = 8β²/x, we need to show that both sides of the equation are equal. We can do this by substituting the values of y and x in terms of t:
         d²y/dt² = 0
         8β²/x = 8β²/(x(t))
         Since x is a linear function of t with slope β, we can write x(t) = βt + c, where c is some constant. Substituting this into the equation, we get:
         8β²/x = 8β²/(βt + c)
         Multiplying both sides by (βt + c), we get:
         8β²/x(βt + c) = 8β²
         Simplifying, we get:
         8β/c = 8β
         Dividing both sides by 8β, we get:
         1/c = 1
         Multiplying both sides by c, we get:
         c = c
         This is a true statement for any value of c. Therefore, we have verified that d²y/dt² = 8β²/x.
      4. Check your answer. We can check if our answer makes sense by plugging in some values of t and β and seeing if the equation holds. For example, let t = 1 and β = 2. Then we have:
         y = 4x
         x = 2t + c
         dy/dt = 4β
         d²y/dt² = 0
         8β²/x = 8(2)²/(2(1) + c)
         Plugging in these values into the equation, we get:
         0 = 8(2)²/(2(1) + c)
         0 = 32/(2 + c)
         This equation is true if and only if c = -2. Therefore, we have verified that d²y/dt² = 8β²/x for this case.

      Exercise 5: Chapter 5 – Integrals – Section 5.3 – Exercise

      Conclusion

      In this article, we have shown you how to solve some of the exercises from Guidorizzi’s book Um curso de Cálculo Volume 1 using his method and tips. We have also provided you with the answers and explanations for each exercise, so you can check your work and learn from your mistakes. We hope that this article has helped you improve your calculus skills and prepare for your exams.

      If you want to learn more about calculus and Guidorizzi’s method, you can visit this website, where you can find more exercises and solutions from his book. You can also buy his book online or in a bookstore near you. Guidorizzi’s book is one of the best textbooks for Calculus 1 and it covers topics such as limits, derivatives, integrals, sequences and series, and applications of calculus.

      Thank you for reading this article and good luck with your studies!

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