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Common Errors in Calculus Exercises

September 13th, 2016

Common Errors in Calculus Exercises

Practical exercises make an important part of a math course, which helps to deeper understand all theoretical information and further study more complex math sections. Quite often students do not even realize that they make mistakes and why that happens. In this article, we will analyze the most common student’s problems met while solving calculus homework.

Not always elementary transformations help us construct a graph of the given function. Often it is easier to do with the help of the derivative, but even this method of the functions properties investigation sometimes causes difficulties and may bring the wrong answer.

A point, which is not in the domain of the given function, is considered as a critical point:

Exercise 1: For given function @$y=\frac{x^2}{x+2}@$ we should find the critical points:

Wrong solution

If @$y=\frac{x^2}{x+2}@$, than @$y’=\frac{2x(x+2)-x^2}{(x+2)^2}=\frac{x^2+4x}{(x+2)^2}@$ and @$-4,-2,0@$ are the critical points.


Please pay attention that when @$x=-2@$, the given function does not exist.

So the correct answer is @$x=-4@$ and @$x=0@$.

The extremum point is considered the point at which the derivative is equal zero.

Exercise 2: Find extremum points of the function @$y = x^3@$.

Wrong solution

For the function @$y = x^3@$, the derivative is @$y’= 3x^2@$. Obviously, the function is equal to 0 when @$x = 0@$.
So, @$x = 0@$ is the extremum point.
Answer: @$x = 0@$.
If you find the point @$x = 0@$ on the coordinate line and determine the derivative signs on one side and the other from that point, it turns out the derivative does not change the sign while going through this point, and therefore, it is not a point of extreme.

Correct answer: the function has no extremal points

Students do not see the difference between extremes and the extremum point.

Find extrema of the function @$y = x^2 + 2x + 3@$.

Wrong solution

For the function @$y = x^2 + 2x + 3@$, find the extremum:
@$y’= 2x + 2@$
@$y’= 0@$ for @$2x + 2 = 0@$ and @$x = -1@$. Verification shows that the sign of the derivative changes from @$-@$ to @$+@$ when passing through the point @$x = -1@$ from the left to the right.
Answer: @$x = -1@$.
In this solution, we found an extreme point, namely the minimum point. It’s necessary to calculate an appropriate value in order to find the extremum (to find the function value at this extremum point).

Correct solution is to continue the above solution

@$x_{min} = -1@$,
@$y_{min} = y (x_{min}) = y (-1) = (-1)^2 + 2 · (-1) + 3 = 1 – 2 + 3 = 2@$.
Answer: @$2@$.

We hope you’ll make use of this information but if any other questions occur, you’re welcome to ask them in comments in our blog or in our Homework Answers section. Be attentive in solving your math homework and follow us!