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Tarina V.

Mechanical Engineering Student

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Spanish

TutorMe

Question:

Maribel tiene hambre. Usa verbos subjuntivo para indicar que hay comida en la nevera.

Tarina V.

Answer:

[Yo] Recomiendo ("recomendar" conjugated to present tense, referring to oneself) que chequees (subjunctive "you" form of "chequear")/mires (subjunctive "you" form of (mirar) en la nevera [porque tenemos comida, or get creative].

Calculus

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Question:

Find the x-values of the local maximums and minimums of f(x) = sin(x)cos(x) between [0, 2(pi)].

Tarina V.

Answer:

To find general extrema (maximums and minimums) of a function, you must first derive it. This is because maximums and minimums occur at points where the tangent line of a function has a slope of 0, or, in the case of cusps, is undefined. In this case, we need to use the product rule since f(x) = sin(x) * cos(x). f'(x) = cos(x)cos(x) + sin(x)(-sin(x)) Now we set this function equal to 0 to find the x-values of the points that have tangents of 0. 0 = cos(x)cos(x) - sin(x)sin(x) 0 = cos^2(x) - sin^2(x) 0 = cos(2x) For which x-values does cos = 0? 2x = (pi)/2, 3(pi)/2, 5(pi)/2... We can simplify all of this values to a general solution: 2x = ((pi) + n*2(pi))/2 where n = 0, 1, 2, 3, .... x = ((pi) + n*2(pi))/4 So we use this general solution to find specific x-values between [0, 2(pi)]. (2(pi) = 8(pi)/4) x = (pi)/4 (n=0), 3(pi)/4 (n=1), 5(pi)/4 (n=2), and 7(pi)/4 (n=3) In this case, the end points of 0 and 2(pi) are included in the range, so these are also local extrema. x = 0, (pi)/4, 3(pi)/4, 5(pi)/4, 7(pi)/4, 2(pi)

Algebra

TutorMe

Question:

What changes need to be made to the function f(x) = x^2 + 2 in order to reflect the function over the x-axis and shift it three units in the positive x-direction?

Tarina V.

Answer:

When transforming polynomial functions one needs to remember: f(x) = a(x - h)^2 + k where a = width and direction of polynomial h = distance from the origin in the x-direction k = distance from the origin in the y-direction The original function put in this format is f(x) = 1(x - 0)^2 + 2 To reflect the function over the x-axis, a needs to switch sign. Right now, a directs the function in the positive direction, and we need to switch it to the negative direction. However, this doesn't reflect the function completely over the x-axis, because the function stays +2 units in the y-direction. We need the function -2 units in the y-direction for a complete reflection over the x-axis, so we multiply the whole function by -1. f(x) = -1 * (1(x - 0)^2 + 2) f(x) = -1(x - 0)^2 - 2 Then, to shift, or translate, the function three units in the positive x-direction, we only need to change the value of h. h = 3 because we are translating three units. This value is positive because we're translating in a positive direction. f(x) = -1(x - 3)^2 - 2 Note that even though we are shifting in the positive direction, the horizontal translation is represented by x - 3. If you look to the first formula, the format I used for translating the polynomials shows x - h. Simplify your answer f(x) = -(x - 3)^2 - 2 OR f(x) = -x^ + 6x - 11

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