Quadratic function with roots in [0, 1]. Prove that f(0)≥49 or f(1)≥49

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Quadratic function with roots in [0, 1]. Prove that f(0)≥49 or f(1)≥49

Chelsea Chen · Accepted Answer

Step 1Notef(12)=(12-a)(12-b)≥136⇒2ab-(a+b)+49≥0Use&amp;nbsp;a+b≥2ab&amp;nbsp;to factorize above inequality as(ab−13)(ab−23)≥0which leads to either&amp;nbsp;ab≤13&amp;nbsp;or&amp;nbsp;ab≥23. Examine the two cases below.Case (1)&amp;nbsp;ab≥23. Evalute&amp;nbsp;f(0)f(0)=(0-a)(0-b)=ab≥49Case (2)&amp;nbsp;ab≤13&amp;nbsp;leads to&amp;nbsp;-ab≥-19&amp;nbsp;Take&amp;nbsp;f′(x)=2x−(a+b)&amp;nbsp;and evalute f(1) in two ways,f(1)=f(12)+∫1/21f&#039;(x)dx=f(12)+34-12(a+b)f(1)=1−(a+b)+abEliminate&amp;nbsp;a+b&amp;nbsp;and apply&amp;nbsp;f(12)≥136,&amp;nbsp;-ab≥-19&amp;nbsp;to obtainf(1)=2f(12)+12-ab≥236+12-19=49⇒f(1)≥49

jmroberts70pbo2 · Accepted Answer

Step 1From&amp;nbsp;f(12)≥136&amp;gt;0&amp;nbsp;it is clear that the roots a and b of f are both either, to the left or to the right of&amp;nbsp;x=12When&amp;nbsp;0≤a≤b≤12, the value (0) is less than f(1) which is easy to see and, by symmetry, it is enough to study this case (in the other case,&amp;nbsp;12≤a≤b≤1&amp;nbsp;we obviously have&amp;nbsp;f(0)&amp;gt;f(1)&amp;nbsp;which it is a very elementary observation).So with the data{f(x)=x2−(a+b)x+ab0≤a≤b≤12f(12)≥136&amp;nbsp;we prove that&amp;nbsp;f(1)≥49First an equivalence whose proof is immediatef(1)≥49⇔ab-a-b+59≥0&amp;nbsp;(1)Also if&amp;nbsp;f(1)≥49&amp;nbsp;this implies by convexity that the point &amp;nbsp;is outside the parabola and consequently there is the tangent of positive slope that passes through&amp;nbsp;(1,49)&amp;nbsp;and touches the parabola at a point of abscissa&amp;nbsp;x0∈[49,1)Now this tangent, after elementary calculation have the following equation:y-49x-1=2-a-b-259+ab-a-bSince this tangent exists if and only if59+ab-a-b≥0we are done (because of the equivalence (1) above).

Quadratic function with roots in \(\displaystyle{\left[{0},\ {1}\right]}\).

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