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# Tutor profile: Chloe M.

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Chloe M.
Three years tutoring experience with a passion for mathematics!
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## Questions

### Subject:Pre-Calculus

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Question:

Simplify the following equation: [(2(a^2)b) / (3(a^4)(b^5)] ^ (-2)

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Chloe M.

To complete this problem, let's revisit a few important exponent rules: 1. x^(-a) = 1 / (x^a) 2. (x^a) * (x^b) = (x^(a+b)) 3. (x^a) / (x^b) = (x^(a-b)) 4. ((x^a)^b) = (x^(a*b)) Now that we've reviewed these rules, we can begin! 1. We want to start from the outside in according to P.E.M.D.A.S. If you don't recall what P.E.M.D.A.S. is, it stands for parenthesis, exponents, multiplication, division, addition, subtraction. It's the step by step way you approach operators in a mathematics problem. Based on this, we're going to address the -2 exponent on the outside of the parenthesis first using rule 1 from above: [(2(a^2)b) / (3(a^4)(b^5)] ^ (-2) = 1 / [(2(a^2)b) / (3(a^4)(b^5)] ^ (2) 2. Since a fraction over a fraction makes things almost as complicated as an expression with a negative exponent, we're going to use our keep, change, flip rule to force the expression into a single fraction. If you don't recall what the keep, change, flip rule is, it's when you keep the top fraction (the 1 in our case), change the division sign to a multiplication sign, and flip the bottom fraction so that the bottom fraction's numerator becomes it's denominator and vice versa. Applying the keep, change, flip rule to our previous expression, we get: 1 / [(2(a^2)b) / (3(a^4)(b^5)] ^ (2) = [(3(a^4)(b^5) / (2(a^2)b)] ^ (2) 3. Now, according to P.E.M.D.A.S., we need to distribute our exponent on the outside of our parenthesis. Using rule 4 from above, we get: [(3(a^4)(b^5) / (2(a^2)b)] ^ (2) = (9(a^6)(b^10)) / (4(a^4)(b^2)) 4. We can take this one step further using rule 3 from above: 9(a^6)(b^10)) / (4(a^4)(b^2)) = (9/4)(a^2)(b^8)

### Subject:Pre-Algebra

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Question:

Find the vertical and horizontal asymptotes of the following equation: f(x) = (5(x^2)) / ((x^2) + x - 2)

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Chloe M.

To complete this problem, we have to understand the definition of a vertical and horizontal asymptote: *A vertical asymptote occurs at values of x for which f(x) is undefined. Examples of this include any values of x that would make the denominator of a function equal to zero or any values of x that would make the value under a square root negative. *A horizontal asymptote occurs at values of y for which the function f(x) approaches, but never reaches. Horizontal asymptotes occur when a function f(x) is a quotient, or in other words, when a function is a fraction. In this circumstance, there are three potential outcomes: 1. If the degree of the numerator is larger than the degree of the denominator, the function f(x) does not have a horizontal asymptote. 2. If the degree of the numerator is equivalent to the degree of the denominator, the function f(x) has a horizontal asymptote at the value of the quotient of the leading coefficient of the numerator over the leading coefficient of the denominator. 3. If the degree of the numerator is less than the degree of the denominator, the function f(x) has a horizontal asymptote at y = 0. Now that we've refreshed our memory, we are ready to determine the horizontal and vertical asymptotes of our function f(x) = (5(x^2)) / ((x^2) + x - 2). 1. Based on our definition, a vertical asymptote occurs when the denominator of our function is equivalent to zero: ((x^2) + x - 2) = 0 2. To compute the solution, we must factor our denominator: ((x^2) + x - 2) = (x + 2)(x - 1) 3. We can now set each factored piece equal to zero to obtain the values at which f(x) has a horizontal asymptote: (x + 2) = 0 (x - 1) = 0 x = -2 x = 1 4. Looking back at our definition, a horizontal asymptote occurs at the value of the quotient of the leading coefficient of the numerator over the leading coefficient of the denominator as the degree of our numerator is equivalent to the degree of our denominator. Thus, we compare the two: (5(x^2)) / (x^2) = 5 5. Putting it all together we see that for the function f(x) = (5(x^2)) / ((x^2) + x - 2), we have two vertical asymptotes at x = -2 and x = 1, and horizontal asymptote at y = 5.

### Subject:Calculus

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Question:

Derive the following equation: f(x) = (sin (x^(1/3))^(1/3)

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Chloe M.

To complete this derivative, we will need to use the chain rule. With chain rule, it's important to remember to work from the outside in. This ensures that you don't forget to derive any part of the equation! 1. Let's let u = x^(1/3). Not only does it make things easier on the eyes, but it also helps us apply the first set of our compounding chain rule. If u = x^(1/3) then our equation becomes (sin(u))^(1/3). 2. Now that we've made the equation a bit simpler, let's apply the chain rule to begin our differentiation process: ***Equation reminder*** Chain rule looks like this: [d/dx] (f(x))^n = n(f(x))^(n-1) * (f'(x)) [d/du] (sin(u))^(1/3) = (1/3) (sin(u))^(-2/3) * cos(u) * u' 3. You might be asking me what that u' is doing at the end of our equation. Remember that we let u = x^(1/3)? We can't forget to derive this part! So, let's apply the chain rule again: u' = [d/dx] x^(1/3) = (1/3) x^(-2/3) * 1 4. Now we can put it all together! Let's plug back in for u and u'. We know that: u = x^(1/3) And that: [d/du] (sin(u))^(1/3) = (1/3) (sin(u))^(-2/3) * cos(u) * u' And that: u' = [d/dx] x^(1/3) = (1/3) x^(-2/3) * 1 So, plugging back in, we get: f'(x) = [d/dx] (sin (x^(1/3))^(1/3) = (1/3) (sin(x^(1/3)))^(-2/3) * cos(x^(1/3)) * (1/3) x^(-2/3) * 1 *Note that we can simplify this by moving terms with negative exponents to the denominator and leaving terms with positive exponents as our numerator. If we chose to simplify, our result would look like this: f'(x) = (cos(x^(1/3)) / [9 * (xsin(x^(1/3)))^(2/3)] *Also note that you do not have to use the u substitution in step 1. The more comfortable you get with chain rule, the less you will find you need to use this trick!

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