#1**0 **

This question appears to be a duplicate. Go to https://web2.0calc.com/questions/domain_60627#r1 to see the explanation for how to determine the domain of this function.

The3MathketeersSep 7, 2023

#1**+1 **

When we are dealing with square roots, the domain of the square root is restricted to when the radicand, also known as the argument of the square root, must be greater than or equal to zero. There are two radicands that contain variables, and we need to find the values where all of them produce values greater than or equal to zero.

The simplest of these to determine the domain is the term \(\sqrt{x - 2}\). This one is relatively straightforward.

\(x - 2 \geq 0 \\ x \geq 2\)

Now, we deal with the harder term, \(\sqrt{(x - 1)(x + 2)(x + 4)}\). Once again, the radicand must be greater than or equal to zero.

\((x - 1)(x + 2)(x + 4) \geq 0 \\ x = 1 \text{ or } x = -2 \text{ or } x = -4 \)

By using the Zero Product Property, we found x-values, x = 1, x = -2, and x = -4 that make the left-hand side of the inequality equal to zero. Now, we investigate the sign of the product to the left and right of these boundary x-values where the value changes from positive to negative or otherwise.

\(x < -4\) | \(-4 < x < -2\) | \(-2 < x < 1\) | \(x > 1\) | |

Sign of \((x - 1)(x + 2)(x + 4)\) | - | + | - | + |

Now that we have created the sign chart, we now know when the radicand is greater than or equal to zero. This means that we know that the restriction on the domain values are \(x \geq 2 \text{ and } (-4 \leq x \leq -2 \text{ or } x \geq 1)\). Put more simply, this just simplifies to \(x \geq 2\).

Written in interval notation, the domain of \(f(x)\) is \([2, \infty)\)

.The3MathketeersSep 3, 2023

#3**+1 **

I read the other answers already provided for this particular problem, and I could not make sense of them, so I will post my own approach to this problem. Before we start with the nitty-gritty details of this problem, we should first determine what possible values of A exist in base 63. In the base 63 system, the smallest possible digit is the digit representing 0 in base 10, and the largest possible digit represents 62 in base 10. Therefore, we conclude that \(0 \leq A \leq 62\). This will become important when we generate values of A.

First, I will expand the base 63 number into its more familiar base 10 form by expansion. However, since we only care about the unit's digit of this particular number, I will simplify as much as possible to make the computation easier.

\(A7894321_{63} = 1 \times 63^0 + 2 \times 63^1 + 3 \times 63^2 + 4 \times 63^3 + 9 \times 63^4 + 8 \times 63^5 + 7 \times 63^6 + A \times 63^7\)

The first observation I made is that the unit's digit of \(63^p\) and \(3^p\) for nonnegative integer powers p have the same unit's digit. This occurs because \(63^p \pmod{10} \equiv 3^p \pmod{10}\). This means I can simplify the nasty expansion to the following. In order to ensure that I am writing mathematically, I will define a new function \(U(x)\) which returns the unit's digit of an integer x.

\(U(A7894321_{63}) = 1 \times 3^0 + 2 \times 3^1 + 3 \times 3^2 + 4 \times 3^3 + 9 \times 3^4 + 8 \times 3^5 + 7 \times 3^6 + A \times 3^7\)

I began exploring the powers of 3, and I immediately noticed a pattern among the powers of 3 and their units digit.

p | 0 | 1 | 2 | 3 | 4 | 5 | 6 | ... |

3^{p} | 1 | 3 | 9 | 27 | 81 | 243 | 729 | ... |

U(3^{p}) | 1 | 3 | 9 | 7 | 1 | 3 | 9 | ... |

Observing the first few values of \(3^p\) should convince yourself that a never-ending pattern exists where 3 raised to the power of 0, 1, 2, and 3 have unique unit's digits, but the pattern repeats thereafter. This essentially means that we can simplify the powers further by applying the rule that \(U\left(3^p\right) = U\left(3^{p \pmod{4}}\right)\)

\(U(A7894321) = 1 \times 3^0 + 2 \times 3^1 + 3 \times 3^2 + 4 \times 3^3 + 9 \times 3^0 + 8 \times 3^1 + 7 \times 3^2 + A \times 3^3\)

Given our observations from the previous investigation of the powers of 3, we can reference that table to determine the unit's digit of the powers of 3.

\(U(A7894321) = 1 \times 1 + 2 \times 3 + 3 \times 9 + 4 \times 7 + 9 \times 1 + 8 \times 3 + 7 \times 9 + A \times 7\)

We can do these multiplications without a calculator unlike before where the calculation would have been unfeasible.

\(U(A7894321) = 1 + 6 +27 + 28 + 9 + 24 + 63 + 7A\)

We could do the addition now, but we can reduce the two-digit numbers to its final digit and do the addition because we only care about the unit's digit.

\(\begin{align*} U(A7894321) &= 1 + 6 + 7 + 8 + 9 + 4 + 3 + 7A \\ &= 38 + 7A \\ &= 8 + 7A \end{align*} \)

We finally have a nice and compact representation of the unit's digit. Of course, we want the unit's digit of this base 63 number to be zero. In order to make \(U(8 + 7A) = 0\). The first part of the addition has a unit's digit of 8. In order to make the sum have a final digit of 0, it would be required that \(U(7A) = 2\).

I will once again try to generate a table of values and observe a pattern

A | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | ... |

7A | 0 | 7 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 70 | 77 | 84 | ... |

U(7A) | 0 | 7 | 4 | 1 | 8 | 5 | 2 | 9 | 6 | 3 | 0 | 7 | 4 | ... |

Based on these observations, \(U(7A) = 2\) when \(U(A) = 6\). In other words, the unit's digit of A needs to be 6. Now, all we need to do is list the values of A with a unit's digit of 6. Also, recall from the beginning that we determined that \(0 \leq A \leq 62\).

This means that \(A = \{6, 16, 26, 36, 46, 56\}\). All values in this set are possible values for A.

The3MathketeersSep 1, 2023

#1**0 **

We can find the maximum and minimum of \(y = \frac{x + 1}{x^2 - x + 1}\) by taking the derivative and determining what x-values make the derivative zero. To be honest, I always have trouble remembering the derivative quotient rule. I have to rehearse "Low D-High minus High D-Low" so that I can recall the derivative quotient rule every time!

\(y = \frac{x + 1}{x^2 - x + 1} \\ \frac{\text{d}y}{\text{d}x} = \frac{(x^2 - x + 1)(1) - (x + 1)(2x - 1)}{(x^2 - x + 1)^2} \\ \frac{\text{d}y}{\text{d}x} = \frac{x^2 - x + 1 - (2x^2 - x + 2x - 1)}{(x^2 - x + 1)^2} \\ \frac{\text{d}y}{\text{d}x} = \frac{x^2 - x + 1 - 2x^2 -x + 1}{(x^2 - x + 1)^2} \\ \frac{\text{d}y}{\text{d}x} = \frac{-x^2 -2x + 2}{(x^2 - x + 1)^2}\)

Now that we have the derivative, we want to identify the critical points, which are the x-values of possible maxima and minima of this particular equation. This can be accomplished by setting the derivative to zero and solving for x.

\(\frac{-x^2 - 2x + 2}{(x^2 -x + 1)^2} = 0 \\ -x^2 -2x + 2 = 0 \\ x^2 + 2x = 2 \\ x^2 + 2x + 1 = 2 + 1 \\ (x + 1)^2 = 3 \\ |x + 1| = \sqrt{3} \\ x_1 = \sqrt{3} - 1 \text{ or } x_2 = -\sqrt{3} - 1\)

We have successfully identified that these x-values are potentials for extrema, but we still have to confirm that these x-values indeed are extrema of this rational function. To do this, we will identify the sign of the first derivative of the x-values to the left and right of the critical points. If the sign changes, then we have identified a maximum or minimum. Since we have found the zeroes of the rational function above, we can factor the numerator. The denominator is a squared quantity, so the quantity will always be positive over the real numbers.

\(\frac{(x-(-\sqrt{3} - 1)(x - (\sqrt{3} - 1))}{(x^2 -x + 1)^2}\)

Left | Zero | Right | Extrema? |

+ | \(x_1 = \sqrt{3} - 1\) | - | Since the sign changes from postiive to negative, this extremum is a maximum. |

- | \(x_2 = -\sqrt{3} - 1\) | + | Since the sign changes from negative to positive, this extremum is a minimum. |

Now that we have confirmed that these critical points are indeed the maximum and minimum of this equation, we can substitute these values into the original equation and find the difference of the y-values.

\(\text{Let } x = \sqrt{3} - 1: \\ y = \frac{\sqrt{3} - 1 + 1}{(\sqrt{3} - 1)^2 - (\sqrt{3} - 1) + 1} \\ y = \frac{\sqrt{3}}{3 - 2\sqrt{3} + 1 - \sqrt{3} + 1 + 1} \\ y = \frac{\sqrt{3}}{-3\sqrt{3}+6}\) | \(\text{Let } x = -\sqrt{3} - 1: \\ y = \frac{-\sqrt{3} - 1 + 1}{(-\sqrt{3} - 1)^2 - (-\sqrt{3} - 1) + 1} \\ y = \frac{-\sqrt{3}}{3+2\sqrt{3} + 1 + \sqrt{3} + 1 + 1} \\ y = \frac{-\sqrt{3}}{3\sqrt{3}+6}\) |

The question asks for the sum of the maximum and the minimum points.

\(\begin{align*} \frac{\sqrt{3}}{6 - 3\sqrt{3}} - \frac{\sqrt{3}}{6 + 3\sqrt{3}} &= \frac{\sqrt{3}}{6 - 3\sqrt{3}} * \frac{6 + 3\sqrt{3}}{6 + 3\sqrt{3}} - \frac{\sqrt{3}}{6 + 3\sqrt{3}} * \frac{6 - 3\sqrt{3}}{6 - 3\sqrt{3}} \\ &= \frac{6\sqrt{3} + 3 * 3}{36 - 9 * 3} - \frac{6\sqrt{3} - 3 * 3}{36 - 9 * 3} \\ &= \frac{6\sqrt{3} - 6 \sqrt{3} + 9 + 9}{9} \\ &= \frac{18}{9} \\ &= 2 \end{align*}\)

I am nearly in disbelief that the sum simplifies so nicely!

The3MathketeersAug 30, 2023