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Question Number 206212 by cortano21 last updated on 09/Apr/24

Commented by Frix last updated on 09/Apr/24

$f^{'} (0) = c which is independent of a$
Commented by TheHoneyCat last updated on 10/Apr/24
Really? So if d=c=b=0, the maximum value of a is 0? and not 1/3? hum...��
Commented by Frix last updated on 10/Apr/24

$I only stated what I stated .$
Commented by TheHoneyCat last updated on 10/Apr/24
Ok, apology. I thought you meant f'(0)=c "is the maximum value of a"... ��
	Answered by TheHoneyCat last updated on 10/Apr/24
	$f(x)=ax^3 +bx^2 +cx+d f ′(x)=3ax^2 +2bx+c is an order two polynomial It has an inflexion at : x_0 :=((−b)/(3a)) if x_0 :=∉ ]0,1[, this function is monotonous. it reaches its extrema at the edges so those mxima are: f ′(0)=c and f ′(1)=3a+2b+c so an equivalent requirement to ∣f ′∣ is: ∣c∣≤1 & ∣3a+2b+c∣≤1 if, however x_0 ∈]0,1[ the function is not monotonous. but it is on [0,x_0 ] and [x_0 ,1] so we get the folowing maxima: f ′(0)=c and f ′(x_0 )=(−/(18))(b^2 /a) and f ′(1)=3a+2b+c so an equivalent condition will be: ∣c∣≤1 & b^2 ≤18∣a∣ &∣3a+2b+c∣≤1 To make things more simple, I will assume a≥0. a<0 is messy especially because it can lead to the maximum beeing a=0... But if you want it, repost the question and I′ll do it. So (summerizing) either way, we have the folowing equivalence: ∣f ′(x)∣≤1 ⇔ ( ((−b)/(3a))∉[0,1] & ∣c∣≤1 & ∣3a+2b+c∣≤1) or (((−b)/(3a))∈[0,1] & ∣c∣≤1 & ∣3a+2b+c∣≤1 & b^2 ≤18∣a∣) ⇔∣c∣≤1 & ∣3a+2b+c∣≤1 &[(((−b)/(3a))∉[0,1]) or (((−b)/(3a))∈[0,1] & b^2 ≤18∣a∣)] ⇔∣c∣≤1 & ∣3a+2b+c∣≤1 &[(((−b)/(3a))∉[0,1]) or b^2 ≤18∣a∣] ⇔∣c∣≤1 & ∣3a+2b+c∣≤1 &[((−b)/(3a))≤0 or 1≤((−b)/(3a)) or b^2 ≤18∣a∣] ⇔∣c∣≤1 & ∣3a+2b+c∣≤1 &[0≤b or (3a+b)≤0 or b^2 ≤18a] This (last line) is the exact requirement on the coeficients to have ∣f ′(x)∣≤1 over [0,1] So we are simply looking for the biggest a verifying these. Notice that the part “∣c∣≤1” just does not involve a... Let us split the cases of the last term. if 0≤b the only constraint on a is: ∣3a+2b+c∣≤1 ⇔3a+2b+c∈[−1,1] ⇔3a∈[−(2b+c+1),1−2b−c] ⇔a∈[−(2b+c+1)/3,(1−2b−c)/3] so if 2b+c<1 then the max value of a will be ((1−2b−c)/3) (if 2b+c>1 we just don′t get an a from this part of the formula) If 3a+b≤0 ⇔a∈[0,−b/3] (so b has to be negtive) reusing previous computations we know: a∈[((−2b−c−1)/3),((−2b−c+1)/3)] if 2b+c<1 we get that the max value of a is Min{−b/3, (−2b−c+1)/3} =((−2b)/3)+Min{b/3,(1−c)/3} Since ∣c∣≤1 in every case, and in the current case b≤0, it is obvious that we get: ((1−2b−c)/3) again. Last but not least b^2 ≤18a does not constraint the maximal value of a. So if it exist in the case of this inequality, the maximum also has to be ((1−2b−c)/3) again. To conclude, in the case of the requirement a>0 the maximal value that can be taken by a while ∣f ′(x)∣≤1 on [0,1] is: a_(Max) := ((1−2b−c)/3) of course, if 2b+c≥1 that will break a>0. This conclude the discussion. _□$
	$f (x) = {a x}^{3} + {b x}^{2} + c x + d$ $f^{'} (x) = 3 {a x}^{2} + 2 b x + c is an order two polynomial$ $It has an inflexion at :$ $x_{0} := \frac{- b}{3 a}$ $if x_{0} :=\notin] 0, 1 [, this function is monotonous .$ $it reaches its extrema at the edges$ $so those mxima are :$ $f^{'} (0) = c and f^{'} (1) = 3 a + 2 b + c$ $so an equivalent requirement to ∣ f^{'} ∣ is :$ $∣ c ∣⩽ 1 & ∣ 3 a + 2 b + c ∣⩽ 1$ $if, however x_{0} \in] 0, 1 [the function is n o t monotonous .$ $but it is on [0, x_{0}] and [x_{0}, 1] so we get the$ $folowing maxima :$ $f^{'} (0) = c and f^{'} (x_{0}) = \frac{-}{18} \frac{b^{2}}{a} and f^{'} (1) = 3 a + 2 b + c$ $so an equivalent condition will be :$ $∣ c ∣⩽ 1 & b^{2} ⩽ 18 ∣ a ∣ & ∣ 3 a + 2 b + c ∣⩽ 1$ $To make things more simple, I will assume a ⩾ 0 .$ $a < 0 is messy especially because it can$ $lead to the maximum beeing a = 0 . . .$ $But if you want it, repost the question and I^{'} ll$ $do it .$ $So (s u m m e r i z i n g) either way, we have the$ $folowing equivalence :$ $∣ f^{'} (x) ∣⩽ 1$ $\Leftrightarrow (\frac{- b}{3 a} \notin [0, 1] & ∣ c ∣⩽ 1 & ∣ 3 a + 2 b + c ∣⩽ 1)$ $or (\frac{- b}{3 a} \in [0, 1] & ∣ c ∣⩽ 1 & ∣ 3 a + 2 b + c ∣⩽ 1 & b^{2} ⩽ 18 ∣ a ∣)$ $\Leftrightarrow∣ c ∣⩽ 1 & ∣ 3 a + 2 b + c ∣⩽ 1 & [(\frac{- b}{3 a} \notin [0, 1]) or (\frac{- b}{3 a} \in [0, 1] & b^{2} ⩽ 18 ∣ a ∣)]$ $\Leftrightarrow∣ c ∣⩽ 1 & ∣ 3 a + 2 b + c ∣⩽ 1 & [(\frac{- b}{3 a} \notin [0, 1]) or b^{2} ⩽ 18 ∣ a ∣]$ $\Leftrightarrow∣ c ∣⩽ 1 & ∣ 3 a + 2 b + c ∣⩽ 1 & [\frac{- b}{3 a} ⩽ 0 or 1 ⩽ \frac{- b}{3 a} or b^{2} ⩽ 18 ∣ a ∣]$ $\Leftrightarrow∣ c ∣⩽ 1 & ∣ 3 a + 2 b + c ∣⩽ 1 & [0 ⩽ b or (3 a + b) ⩽ 0 or b^{2} ⩽ 18 a]$ $This (last line) is the exact requirement$ $on the coeficients to have ∣ f^{'} (x) ∣⩽ 1 over [0, 1]$ $So we are simply looking for the biggest a verifying$ $these .$ $Notice that the part ‘ ‘ ∣ c ∣⩽ 1^{″} just does not$ $involve a . . .$ $Let us split the cases of the last term .$ $if 0 ⩽ b the only constraint on a is :$ $∣ 3 a + 2 b + c ∣⩽ 1$ $\Leftrightarrow 3 a + 2 b + c \in [- 1, 1]$ $\Leftrightarrow 3 a \in [- (2 b + c + 1), 1 - 2 b - c]$ $\Leftrightarrow a \in [- (2 b + c + 1) / 3, (1 - 2 b - c) / 3]$ $so if 2 b + c < 1 then the \max$ $value of a will be \frac{1 - 2 b - c}{3}$ $(if 2 b + c > 1 we just {don}^{'} t get an a from this$ $part of the formula)$ $If 3 a + b ⩽ 0 \Leftrightarrow a \in [0, - b / 3]$ $(so b has to be negtive)$ $reusing previous computations we know :$ $a \in [\frac{- 2 b - c - 1}{3}, \frac{- 2 b - c + 1}{3}]$ $if 2 b + c < 1 we get that the \max value of$ $a is Min {- b / 3, (- 2 b - c + 1) / 3}$ $= \frac{- 2 b}{3} + Min {b / 3, (1 - c) / 3}$ $Since ∣ c ∣⩽ 1 in every case, and in the current$ $case b ⩽ 0, it is obvious that we get :$ $\frac{1 - 2 b - c}{3} again .$ $Last but not least b^{2} ⩽ 18 a does not constraint$ $the maximal value of a . So if it exist in the case of$ $this inequality, the maximum also has to be$ $\frac{1 - 2 b - c}{3} again .$ $To conclude, in the case of the requirement$ $a > 0 the maximal value that can be taken by$ $a while ∣ f^{'} (x) ∣⩽ 1 on [0, 1] is :$ $a_{Max} := \frac{1 - 2 b - c}{3}$ $of course, if 2 b + c ⩾ 1 that will break a > 0 .$ $This conclude the discussion ._{◻}$
	Answered by Frix last updated on 10/Apr/24

	$f^{'} (x) = g (x) = 3 {a x}^{2} + 2 b x + c$ $a \to \max \Rightarrow a > 0 \Rightarrow$ $g (x) has an absolute \min at x_{0} = - \frac{b}{3 a}$ $Case 1 : - \frac{b}{3 a} \notin] 0, 1 [$ $To get the steepest possible parabola let$ $g (0) = - 1 \land g (1) = 1 (_{the same a}^{g (0) = 1 \land g (1) = - 1 leads to})$ $\Rightarrow$ $b = 1 - \frac{3 a}{2} \land c = - 1 \Rightarrow x_{0} = \frac{1}{2} - \frac{1}{3 a}$ $x_{0} ⩽ 0 \lor 1 ⩽ x_{0} \Rightarrow a ⩽ \frac{2}{3}$ $witb a = \frac{2}{3}$ $g (x) = f^{'} (x) = 2 x^{2} - 1 (or 2 x^{2} - 4 x + 1)$ $Case 2 : - \frac{b}{3 a} \in] 0, 1 [$ $a > 0 \Rightarrow b < 0$ $g (- \frac{b}{3 a}) = - 1 \land g (1) = 1 ()$ $\Rightarrow$ $b = \sqrt{6 a} - 3 a \land c = 3 a - 2 \sqrt{6 a} + 1 \Rightarrow x_{0} = 1 - \frac{\sqrt{2}}{\sqrt{3 a}}$ $0 < x_{0} < 1 \Rightarrow a > \frac{2}{3}$ $g (1) = 1 \Rightarrow - 1 ⩽ g (0) ⩽ 1$ $g (0) = c \Rightarrow - 1 ⩽ 3 a - 2 \sqrt{6 a} + 1 ⩽ 1$ $\Rightarrow 0 ⩽ a ⩽ \frac{8}{3}$ $\Rightarrow a_{\max} = \frac{8}{3}$ $g (x) = f^{'} (x) = 8 x^{2} - 8 x + 1$
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