Part (a) h(1) = f(g(1)) - 6 h(3) = f(g(3)) - 6 h(1) = f(2) - 6

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Presentation transcript:

Part (a) h(1) = f(g(1)) - 6 h(3) = f(g(3)) - 6 h(1) = f(2) - 6 h(3) = f(4) - 6 h(1) = 9 - 6 h(3) = -1 - 6 h(1) = 3 h(3) = -7 Since f(x) & g(x) are differentiable functions, they must also be continuous. Therefore, h(x) is continuous as well. If h(1) = 3 and h(3) = -7, then the Intermediate Value Theorem states that h(x) must be -5 somewhere between x=1 & x=3.

Part (b) From Part (a), we’ve seen that h(x) passes through the points (1,3) and (3,-7). If we connect these two points with a line and find its slope, it will be m=-5. The Mean Value Theorem (MVT) states that there must be at least one value “c” on the interval 1 < c < 3 such that h’(c) = -5

This is the Fundamental Theorem of Calculus (FTC) !!! Part (c) This is the Fundamental Theorem of Calculus (FTC) !!! w’(3) = f(g(3)) * g’(3) - f(1) * 0 w’(3) = f(4) * 2 w’(3) = -1 * 2 w’(3) = -2

Therefore, the slope of g-1(x) @ (2,1) is 1/5. Part (d) Therefore, the slope of g-1(x) @ (2,1) is 1/5. The slope of g(x) @ (1,2) was 5/1. g(x) passes through the following points: (1,2) (2,3) (3,4) (4,6) Therefore, g-1(x) will pass through these points: (2,1) (3,2) (4,3) (6,4) (y-1) = m (x-2) 1/5 or y = 1/5 x + 3/5