tangent line: (y - ) = (x - ) e2

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Presentation transcript:

tangent line: (y - ) = (x - ) e2 f(x) = ln x x f’(x) = 1 - ln x x2 Part (a) mtan = 1 - ln e2 (e2)2 1 - 2 e4 = -1 e4 = f(e2) = ln e2 e2 2 e2 = tangent line: (y - ) = (x - ) 2 e2 -1 e4 e2

Part (b) = 0 1 – ln x = 0 ln x = 1 x = e ln x 1 - ln x f’(x) = f(x) = Multiply both sides by x2. 1 - ln x x2 = 0 1 – ln x = 0 ln x = 1 First, test a value less than e. x=1 is a convenient choice. x = e f’(1) = 1 - ln 1 12 = 1 e Since f’(x) is positive, the graph of f(x) is increasing on this interval.

Part (b) f(x) = ln x x f’(x) = 1 - ln x x2 Next, test a value greater than e. x = e2, x = e3, etc. would be convenient choices. f’(e2) = 1 - ln e2 (e2)2 = e4 1 - 2 = e4 -1 Since f’(x) is negative, the graph of f(x) is decreasing on this interval. e There is a RELATIVE MAX at x = e, since f(x) goes from increasing to decreasing there.

Use the Product Rule to find the 2nd derivative. f(x) = ln x x f’(x) = 1 - ln x x2 Part (c) v u f’(x) = 1 - ln x x2 Use the Product Rule to find the 2nd derivative. f”(x) = (-1/x)(x2) – (1 - ln x)(2x) x4 Set f”(x) = 0. x = 0 or 2 ln x = 3 -x – 2x + 2x ln x = 0 (not in the domain) -3x + 2x ln x = 0 ln x = 3/2 x (2 ln x – 3) = 0 x = e3/2

Part (d) lim - or DNE ln x 1 - ln x f’(x) = f(x) = x x2 (approaching -) (positive,approaching 0) f(x) = ln (0.0001) 0.0001 lim x0+ - or DNE