Finding all a such that f(x)=sin⁡2x−8(a+1)sin⁡x+(4a2+8a−14)x is increasing and has no critical pointsObviously, the first thing I did was to find the derivative of this function and simplify it a bit and I got:f′(x)=4(cos2x−2(a+1)cos⁡x+(a2+2a−4))But now how do I proceed further, had it been a simple quadratic in x.

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Cheryl King · Accepted Answer

f′(x)=4(cos2x−2(a+1)cos⁡x+(a2+2a−4))We need to find a for which f(x) is increasing, which means f&#039;(x)&amp;gt;0.Now,4(cos2x−2(a+1)cos⁡x+(a2+2a−4))&amp;gt;0cos2x−2(a+1)cos⁡x+(a2+2a−4)&amp;gt;0(cos2x−2(a+1)cos⁡x+a2+2a+1)&amp;gt;5((cos⁡x−(a+1))2&amp;gt;5cos⁡x−(a+1)&amp;gt;5  or  cos⁡x−(a+1)&amp;lt;−5a&amp;lt;cos⁡x−1−5ora&amp;gt;cos⁡x−1+5Hence, a&amp;lt;−2−5  or  a&amp;gt;5

kalupunangh · Accepted Answer

Since I got the answer: To solve: f′(x)&amp;gt;0 i.e. f′(x)=4(cos2x−2(a+1)cos⁡x+(a2+2a−4))&amp;gt;0 Completing the square: 4([cos⁡x−(a+1)]2−5)&amp;gt;0 Solving this, considering the critical cases according to what value cosx takes and final answer is: a∈(−∞,−2−5)∪(5,∞)

karton · Accepted Answer

A preliminary exploration indicates that the answer is a∈(−∞,p)∪(q,∞), where p≈−4.25,q≈2.25Here&#039;s the full solution. We want:∀x∈R  f′(x)&amp;gt;0Let y=cos⁡x. Then we want:∀y∈[−1,1]    y2−2(a+1)y+(a2+2a−4)&amp;gt;0This function in y opens upwards (y2 has positive coefficient) and has two x−intercepts (discriminant is 5&amp;gt;0). So, we want either its right x−intercept to be less than -1 or its left x−intercept to be greater than 1:a+1+5&amp;lt;−1   or   a+1−5&amp;gt;1a∈(−∞,−2−5)∪(5,∞)

Finding all a such that f(x)=\sin 2x-8(a+1)\sin x+(4a^2+8

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