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Solution we complete the square to write the

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Solution: We complete the square to write the quadratic function f ( x ) in ver- tex form. f ( x ) = - 3 x 2 + 6 x + 2 = - 3[ x 2 - 2 x ] + 2 = - 3[( x - 1) 2 - ( - 1) 2 ] + 2 = - 3[( x - 1) 2 - 1] + 2 = - 3( x - 1) 2 + 3 + 2 = - 3( x - 1) 2 + 5 . So the vertex of the graph of f is (1 , 5)
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Math 171: Exam 1 (b) [2 points] Does f have a minimum or maximum value? If so, what is it? Solution: Since the coefficient of x 2 is negative, we know that f has a maximum value and no minimum value. The maximum value occurs at the vertex, so it is f (1) = 5. 4. [10 points] Find two numbers whose sum equals 6 and whose product equals 3. Solution: We want two numbers a and b such that: a + b = 6, and ab = 3 . Solving the first equation for b gives us b = 6 - a . Substituting into the second equation, we get a (6 - a ) = 3. Rearranging gives us - a 2 + 6 a - 3 = 0. Using the quadratic formula to solve for a , we have a = - 6 ± p 6 2 - 4( - 1)( - 3) 2( - 1) = - 6 ± 24 - 2 = 3 6 . Taking a = 3 + 6 we get b = 6 - a = 6 - (3 + 6) = 3 - 6. So the two numbers are 3 + 6 and 3 - 6. It is easy to check that these numbers work. 5. Let f ( x ) = x 1 / 3 + 5 2 x 1 / 3 - 1 . (a) [7 points] Find a formula for f - 1 ( x ). Solution: We set y equal to f ( x ) and solve for x : y = x 1 / 3 + 5 2 x 1 / 3 - 1 y (2 x 1 / 3 - 1) = x 1 / 3 + 5 2 yx 1 / 3 - y = x 1 / 3 + 5 2 yx 1 / 3 - x 1 / 3 = y + 5 x 1 / 3 (2 y - 1) = y + 5 x 1 / 3 = y + 5 2 y - 1 x = ( x 1 / 3 ) 3 = y + 5 2 y - 1 3 . Page 2
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Math 171: Exam 1 So f - 1 ( x ) = x + 5 2 x - 1 3 .
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