MATA30H3 Midterm: MATA30 Midterm 2011 Solution
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MATA30H3 Full Course Notes
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0 < ex < thus 0 < 1/ex = e x < and 3 < 2e x + 3 < . Thus the range of f is the interval (3, ). (b) [ 1 mark] show that f is invertible. The graph of f is the graph of y = ex ipped about the y axis, stretched vertically by a factor of 2 and then shifted upward by 3 units. Thus the graph of f is decreasing in its domain. Thus f is a one to one function and so f is invertible. 10 (c) [ 3 marks] find f 1, the inverse of f , and give its domain and range. solution : y = 2e x + 3 y 3 = 2e x (y 3) 2 (cid:19) = x thus the inverse of f is f 1(x) = ln(cid:18) x 3. 2 (cid:19) = [ln (x 3) ln(2)] = ln(2) ln (x 3)
