0.1-0.4 - 016A S2 Homework 1 Solution Jae-young Park...

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016A S2 Homework 1 Solution Jae-young Park * September 2, 2007 0.1 *10 Use intervals to describe the real numbers satisfying the in- equality x ≥ - 1 and x < 8. Solution You may change the expression little bit: - 1 x < 8, which is equivalent. Using interval notation, we can describe this inequality as [ - 1 , 8) 0.1 *20 If f ( x ) = x 2 + 4 x + 3, find f ( a - 1) and f ( a - 2). Solution To find f ( a - 1), we substitute a - 1 for every x appears in the formula defining f ( x ). So we get, f ( a - 1) = ( a - 1) 2 + 4( a - 1) + 3 This expression can be simplified using ( a - 1) 2 = a 2 - 2 a + 1. So we get f ( a - 1) = [ a 2 - 2 a + 1] + [4 a - 4] + 3 = a 2 + 2 a Similarly for f ( a - 2). Substituting a - 2 for x , we get f ( a - 2) = ( a - 2) 2 + 4( a - 2) + 3 Using the fact that ( a - 2) 2 = a 2 - 4 a + 4, simplify this expression to get f ( a - 2) = [ a 2 - 4 a + 4] + [4 a - 8] + 3 = a 2 - 1 * jaypark at m a t h . b e r k e l e y . e d u. GSI for 16A 205,211,213 1
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0.1 *22 (a) Suppose that b = 20. Find the response of the muscles when x = 60. Solution Since b = 20, the response of the muscle R ( x ) is given by R ( x ) = 100 x 20 + x , for x 0 Substituting x = 60, we find R (60). R (60) = 100 · 60 20 + 60 = 100 · 60 80 = 6000 80 = 600 8 = 75 So the answer is 75%. 0.1 *22 (b) Determine the value of b if R (50) = 60 -that is, if a concentration of x = 50 units produces a 60% response. Solution Substituting 50 for x , we get R (50) = 100 · 50 b + 50 Since R (50) = 60, we get 100 · 50 b + 50 = 60 Multiplying both sides by b + 50, we get an equation 60( b + 50) = 100 · 50 = 5000 = 60 b + 3000 = 5000 Subtracting both sides by 3000, we get 60 b = 2000 = b = 2000 60 = 100 3 You may verify this answer by plugging b = 100 3 in R ( x ) and finding the value of R (50). So the answer is b = 100 3 0.1 *25 Describe the domain of the function g ( x ) = 1 3 - x Solution Implicitly, we are working with real numbers. Also, recall that we understand the intended domain to consist of all numbers for which the defining formula make sense. Since 3 - x is defined if and only if 3 - x 0 and dividing by zero doesn’t make sense, the defining formula makes sense if and only if 3 - x > 0. So the domain is x > 3. 2
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0.1 *26 Describe the domain of the function g ( x ) = 4 x ( x +2) Solution As we did in the previous problem, we find the values which the defining formula makes sense. The formula for g ( x ) makes sense unless the denominator x ( x + 2) is zero. Finding the zero of x ( x + 2), we get x = 0 or x = - 2. So the domain is all real numbers except x = 0 , - 2. You may express this domain by R r { 0 , - 2 } . 0.1 *30 Decide which curves are graphs of functions. Solution Look at the figure in you book, page 15. Apply the vertical line test. You can easily find some vertical line which touches the curve at two points. Therefor it is not a graph of a function. 0.1 *56 Compute f (1) , f (2) , f (3) for the following funtion f ( x ) = 3 / (4 - x ) for x < 2 2 x for 2 x < 3 x 2 - 5 for 3 x Solution For x = 1, since x = 1 < 2, we get f (1) = 3 4 - 1 = 1. For x = 2, since 2 x = 2 < 3, we have f (2) = 2 · 2 = 4. For x = 3, since 3 x , f (3) = 3 2 - 5 = 4 = 2 0.1 bonus *50 Is the point (2 / 3 , 5 / 3) on the graph of the function g ( x ) = ( x 2 + 4) / ( x + 2)?
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