Let f(x) = 3tan^4x+2k and g(x) = -tan^4x+8ktan^2 x + k for 0 ≤ x ≤ 1, where 0 < k < 1. The graphs of f and g intersect at exactly one point. Find k. My book gets an answer of 1/4 where as I get k=4. Please explain.
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Expert's answer
2022-01-11T17:39:30-0500
Suppose f(x)=g(x) then we have the following 3tan4x+2k=−tan4x+8ktan2x+k⟹4tan4x=k(8tan2x−1)⟹k=8tan2x−14tan4xBy futher simplification we have the following 4tan4x−8ktan2x+k=0set y=2tan2x to obtain the following y2−4ky+k=0⟹y=24k−±16k2−4k=24k±24k2−k=2k±k(4k−1)for the value of y to be valid we must have k(4k−1)≥0 since 0<k<1 we have 4k≥1⟹k≥41Since the minimum value for k is 41 then it suffices to take k=41 since we have that f and g intersect at a point.
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