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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: ktrah on May 15, 2012, 05:39:07 pm
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Consider the function f: R--> R, f(x) = (x-1)² (x-2) + 1.
Find the real values of h for which only one of the solutions of the equation f(x+h) = 1 is positive.
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The graph of the cubic f(x) has a local maximum @ (1,1) and minimum at approx (1.67, 0.85). It passes through the point (2, 1)
There are only two places (solutions) where f(x) = 1 . ie at x=1 and x=2. Both are positive values and are 1 unit apart.
If h is positive, the graph of f(x+h) is a translation of f(x) in the negative x direction of h units.
Since no vertical translation is involved, f(x+h) = 1 will occur at two places (solutions) also, and again 1 unit apart.
To have only one of these as a positive solution, the other must be zero or negative.
By translating f(x) by 1 unit in the negative x direction, you will get one solution at x=0 and the other as a positive solution at x=1
Therefore, the least value of h must be 1.
If h=2, the two solutions of f(x+h) will occur at x=-1 and x=0; neither of which is positive. Hence h<2
Therefore the domain for h is [1, 2)