math review 2

math review 2 - a. y = 12x 1 2 – 6 x 1 2 x 2 . b. Q = 6K...

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University of British Columbia Sauder School of Business COMM 295 Math Refresher (This assignment will not be graded but is very important to do well in this course) 1. Solve for x: a. 10x – 40 = 2x. b. 10x 2 = 60x 2. Find the derivative of the following functions: a. y = 25. b. y = 12x 3 – 6x 3. Find the slope of the function: y = 12x 2 – 6x at x = 2 and at x = 0. 4. Find the value of x at which the value of y = x 2 – 6x is maximum or minimum. 5. Find the first derivatives of the following functions (w.r.t. x): (a) y = (5x + 2) 2 (use chain rule, don’t use (5x + 2) 2 = 25x 2 +20x +4) (b) y = 5x 2 - 3x. 6. Find the partial Derivatives:
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Unformatted text preview: a. y = 12x 1 2 – 6 x 1 2 x 2 . b. Q = 6K 2 L. 7. Solve the following system of equations: 5Q + 4P = 20 4Q – P = 10. 8. Suppose U(w) = w 0.5 . If U(x – 2) = 20, find the value of x. 9. Suppose y(K, L) = K 2 L. a. Find the partial derivative of y w.r.t K and L (that is, ∂ y/ ∂ K =? and ∂ y/ ∂ L = ?). b. Is ∂ y/ ∂ K increasing or decreasing in K? c. What can you say about the relationship between ∂ y/ ∂ L and L? d. Draw the graphs of ∂ y/ ∂ K vs. K with L fixed at 10. e. Draw the graphs of ∂ y/ ∂ L vs. L with K= 5....
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This note was uploaded on 01/03/2011 for the course COMM 290 taught by Professor Brian during the Winter '09 term at UBC.

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