MaxWong

Also, sometimes it incorrectly determines that a nonexistent limit exists.​

The step-by-step solution is not correct for some functions. For example, check this case: ​

Generally, for piecewise functions, you break the integral at the boundary points of the cases. In this problem, you break the integral at x = 3.

\(\newcommand{\dint}{\displaystyle\int} \dint_0^5 f(x)\,dx = \left(\dint_0^3 + \dint_3^5\right)f(x)\,dx = \dint_0^3 f(x) \,dx + \dint_3^5 f(x)\,dx\)

Now, you can use the fact that f(x) = 3 for 0 < x < 3 and f(x) = x for 3 < x < 5 to continue. Use the power rule of integration.

Moving r to the other side gives \(\left\lfloor\dfrac r4\right\rfloor = 15.5 -r \).

Then 15.5 - r is an integer. Then r = k + 0.5 for some integer k.

Then \(\left\lfloor\dfrac {k + \frac12}4\right\rfloor = 15 - k\).

That means \(15-k\leq \dfrac{k + \frac12}4 < 16 - k\) by definition of floor function. Now it becomes a compound inequality. To proceed, solve the compound inequality and list its integer solutions.

Consider the term in x.

\(7(tx) + (-6x)(10) = -39x\\ (7t - 60)x = 39x\\ 7t - 60 = 39\)

Please continue to solve for the value of t on your own.

\(\begin{array}{cc} &A + 2 = C + 2\\ \implies & A = C \end{array}\)

\(\begin{array}{cc} &B - 2 = D - 2\\ \implies & B = D \end{array}\)

\(\begin{array}{cc} &A + 2 = B - 2\\ \implies & B = A + 4 \end{array}\)

\(\begin{cases} A = C\\ B = D = A + 4 \end{cases}\)

Since A + B + C + D = 36 and with the above system of equation,

\(A + (A + 4) + A + (A + 4) = 36\)

Now you can solve for the value of A. If you know the value of A, you know the values of B, C, D. Then you will be able to find the product ABCD.

This method is nicer and is easier to do! 

So, the graph looks like this:

This graph can be obtained from a graphical calculator.

Can you plot the graph on your own?

Just expand it using tabular method: 

You list the terms of the first polynomial in the columns, and the terms of the second polynomial in the rows:

\(\begin{array}{|c|c|c|} \hline &2y&-1\\ \hline 4y^{10}&&\\\hline +y^9&&\\\hline +y^8&&\\\hline +y^7&&\\\hline \end{array}\)

And then fill in the table with the product of row and column. For example, 4y^{10} * 2y = 8y^{11}, so:

\(\begin{array}{|c|c|c|} \hline &2y&-1\\ \hline 4y^{10}&\color{blue}8y^{11}&\\\hline +y^9&&\\\hline +y^8&&\\\hline +y^7&&\\\hline \end{array}\)

Filling in the table like so:
 

\(\begin{array}{|c|c|c|} \hline &2y&-1\\ \hline 4y^{10}&\color{blue}8y^{11}&\color{blue}-4y^{10}\\\hline +y^9&\color{blue}2y^{10}&\color{blue}-y^9\\\hline +y^8&\color{blue}2y^9&\color{blue}-y^8\\\hline +y^7&\color{blue}2y^8&\color{blue}-y^7\\\hline \end{array}\)

To get the product, add all the blue terms in the table:

\((2y - 1)(4y^{10} + y^9 + y^8 + y^7) = 8y^{11} - 4y^{10} + 2y^{10} - y^9 + 2y^9 - y^8 + 2y^8 - y^7\)

Please do the final simplification on your own.

We pick one term from each polynomial to try to "make" x^2.

\((-5x^3 \color{purple}- 5x^2 \color{blue}- 7x \color{green}+ 1\color{black})(\color{green}x^2 \color{blue}- 6x\color{purple} + 7\color{black})\)

If we combine the colored terms with the same color together, we get terms in x^2. We add them up to get the total:

Term in x^2 = \(\color{purple}(-5x^2)(+7)\color{black}+\color{blue}(-7x)(-6x)\color{black}+\color{green}(+1)(x^2)\color{black}=8x^2\)

Therefore the coefficient of x^2 in the product is 8.