Question Number 206212 by cortano21 last updated on 09/Apr/24
Commented by Frix last updated on 09/Apr/24
$${f}'\left(\mathrm{0}\right)={c}\:\mathrm{which}\:\mathrm{is}\:\mathrm{independent}\:\mathrm{of}\:{a} \\ $$
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
$$\mathrm{I}\:\mathrm{only}\:\mathrm{stated}\:\mathrm{what}\:\mathrm{I}\:\mathrm{stated}. \\ $$
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. _□](https://www.tinkutara.com/question/Q206257.png)
$${f}\left({x}\right)={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$${f}\:'\left({x}\right)=\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{2}{bx}+{c}\:\mathrm{is}\:\mathrm{an}\:\mathrm{order}\:\mathrm{two}\:\mathrm{polynomial} \\ $$$$\mathrm{It}\:\mathrm{has}\:\mathrm{an}\:\mathrm{inflexion}\:\mathrm{at}\:: \\ $$$${x}_{\mathrm{0}} :=\frac{−{b}}{\mathrm{3}{a}} \\ $$$$ \\ $$$$\left.\mathrm{if}\:{x}_{\mathrm{0}} :=\notin\:\right]\mathrm{0},\mathrm{1}\left[,\:\mathrm{this}\:\mathrm{function}\:\mathrm{is}\:\mathrm{monotonous}.\right. \\ $$$$\mathrm{it}\:\mathrm{reaches}\:\mathrm{its}\:\mathrm{extrema}\:\mathrm{at}\:\mathrm{the}\:\mathrm{edges} \\ $$$$\mathrm{so}\:\mathrm{those}\:\mathrm{mxima}\:\mathrm{are}: \\ $$$${f}\:'\left(\mathrm{0}\right)={c}\:\:\:\mathrm{and}\:\:{f}\:'\left(\mathrm{1}\right)=\mathrm{3}{a}+\mathrm{2}{b}+{c} \\ $$$$\mathrm{so}\:\mathrm{an}\:\mathrm{equivalent}\:\mathrm{requirement}\:\mathrm{to}\:\mid{f}\:'\mid\:\mathrm{is}: \\ $$$$\mid{c}\mid\leqslant\mathrm{1}\:\&\:\mid\mathrm{3}{a}+\mathrm{2}{b}+{c}\mid\leqslant\mathrm{1} \\ $$$$ \\ $$$$\left.\mathrm{if},\:\mathrm{however}\:{x}_{\mathrm{0}} \in\right]\mathrm{0},\mathrm{1}\left[\:\mathrm{the}\:\mathrm{function}\:\mathrm{is}\:{not}\:\mathrm{monotonous}.\right. \\ $$$$\mathrm{but}\:\mathrm{it}\:\mathrm{is}\:\mathrm{on}\:\left[\mathrm{0},{x}_{\mathrm{0}} \right]\:\mathrm{and}\:\left[{x}_{\mathrm{0}} ,\mathrm{1}\right]\:\mathrm{so}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\: \\ $$$$\mathrm{folowing}\:\mathrm{maxima}: \\ $$$${f}\:'\left(\mathrm{0}\right)={c}\:\:\mathrm{and}\:{f}\:'\left({x}_{\mathrm{0}} \right)=\frac{−}{\mathrm{18}}\frac{{b}^{\mathrm{2}} }{{a}}\:\:\mathrm{and}\:{f}\:'\left(\mathrm{1}\right)=\mathrm{3}{a}+\mathrm{2}{b}+{c} \\ $$$$\mathrm{so}\:\mathrm{an}\:\mathrm{equivalent}\:\mathrm{condition}\:\mathrm{will}\:\mathrm{be}: \\ $$$$\mid{c}\mid\leqslant\mathrm{1}\:\&\:{b}^{\mathrm{2}} \leqslant\mathrm{18}\mid{a}\mid\:\&\mid\mathrm{3}{a}+\mathrm{2}{b}+{c}\mid\leqslant\mathrm{1} \\ $$$$ \\ $$$$\mathrm{To}\:\mathrm{make}\:\mathrm{things}\:\mathrm{more}\:\mathrm{simple},\:\mathrm{I}\:\mathrm{will}\:\mathrm{assume}\:{a}\geqslant\mathrm{0}. \\ $$$${a}<\mathrm{0}\:\mathrm{is}\:\mathrm{messy}\:\mathrm{especially}\:\mathrm{because}\:\mathrm{it}\:\mathrm{can} \\ $$$$\mathrm{lead}\:\mathrm{to}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{beeing}\:{a}=\mathrm{0}… \\ $$$$\mathrm{But}\:\mathrm{if}\:\mathrm{you}\:\mathrm{want}\:\mathrm{it},\:\mathrm{repost}\:\mathrm{the}\:\mathrm{question}\:\mathrm{and}\:\mathrm{I}'\mathrm{ll} \\ $$$$\mathrm{do}\:\mathrm{it}. \\ $$$$\mathrm{So}\:\left({summerizing}\right)\:\mathrm{either}\:\mathrm{way},\:\mathrm{we}\:\mathrm{have}\:\mathrm{the} \\ $$$$\mathrm{folowing}\:\mathrm{equivalence}: \\ $$$$\mid{f}\:'\left({x}\right)\mid\leqslant\mathrm{1} \\ $$$$\Leftrightarrow\:\left(\:\frac{−{b}}{\mathrm{3}{a}}\notin\left[\mathrm{0},\mathrm{1}\right]\:\&\:\mid{c}\mid\leqslant\mathrm{1}\:\&\:\mid\mathrm{3}{a}+\mathrm{2}{b}+{c}\mid\leqslant\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{or}\:\left(\frac{−{b}}{\mathrm{3}{a}}\in\left[\mathrm{0},\mathrm{1}\right]\:\&\:\mid{c}\mid\leqslant\mathrm{1}\:\&\:\mid\mathrm{3}{a}+\mathrm{2}{b}+{c}\mid\leqslant\mathrm{1}\:\&\:{b}^{\mathrm{2}} \leqslant\mathrm{18}\mid{a}\mid\right) \\ $$$$\Leftrightarrow\mid{c}\mid\leqslant\mathrm{1}\:\&\:\mid\mathrm{3}{a}+\mathrm{2}{b}+{c}\mid\leqslant\mathrm{1}\:\&\left[\left(\frac{−{b}}{\mathrm{3}{a}}\notin\left[\mathrm{0},\mathrm{1}\right]\right)\:\mathrm{or}\:\left(\frac{−{b}}{\mathrm{3}{a}}\in\left[\mathrm{0},\mathrm{1}\right]\:\&\:{b}^{\mathrm{2}} \leqslant\mathrm{18}\mid{a}\mid\right)\right] \\ $$$$\Leftrightarrow\mid{c}\mid\leqslant\mathrm{1}\:\&\:\mid\mathrm{3}{a}+\mathrm{2}{b}+{c}\mid\leqslant\mathrm{1}\:\&\left[\left(\frac{−{b}}{\mathrm{3}{a}}\notin\left[\mathrm{0},\mathrm{1}\right]\right)\:\mathrm{or}\:{b}^{\mathrm{2}} \leqslant\mathrm{18}\mid{a}\mid\right] \\ $$$$\Leftrightarrow\mid{c}\mid\leqslant\mathrm{1}\:\&\:\mid\mathrm{3}{a}+\mathrm{2}{b}+{c}\mid\leqslant\mathrm{1}\:\&\left[\frac{−{b}}{\mathrm{3}{a}}\leqslant\mathrm{0}\:\mathrm{or}\:\mathrm{1}\leqslant\frac{−{b}}{\mathrm{3}{a}}\:\mathrm{or}\:{b}^{\mathrm{2}} \leqslant\mathrm{18}\mid{a}\mid\right] \\ $$$$\Leftrightarrow\mid{c}\mid\leqslant\mathrm{1}\:\&\:\mid\mathrm{3}{a}+\mathrm{2}{b}+{c}\mid\leqslant\mathrm{1}\:\&\left[\mathrm{0}\leqslant{b}\:\mathrm{or}\:\left(\mathrm{3}{a}+{b}\right)\leqslant\mathrm{0}\:\mathrm{or}\:{b}^{\mathrm{2}} \leqslant\mathrm{18}{a}\right] \\ $$$$ \\ $$$$\mathrm{This}\:\left(\mathrm{last}\:\mathrm{line}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{requirement} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{coeficients}\:\mathrm{to}\:\mathrm{have}\:\mid{f}\:'\left({x}\right)\mid\leqslant\mathrm{1}\:\mathrm{over}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{So}\:\mathrm{we}\:\mathrm{are}\:\mathrm{simply}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{the}\:\mathrm{biggest}\:{a}\:\mathrm{verifying} \\ $$$$\mathrm{these}. \\ $$$$\mathrm{Notice}\:\mathrm{that}\:\mathrm{the}\:\mathrm{part}\:“\mid{c}\mid\leqslant\mathrm{1}''\:\mathrm{just}\:\mathrm{does}\:\mathrm{not} \\ $$$$\mathrm{involve}\:{a}… \\ $$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{split}\:\mathrm{the}\:\mathrm{cases}\:\mathrm{of}\:\mathrm{the}\:\mathrm{last}\:\mathrm{term}. \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{0}\leqslant{b}\:\mathrm{the}\:\mathrm{only}\:\mathrm{constraint}\:\mathrm{on}\:{a}\:\mathrm{is}: \\ $$$$\mid\mathrm{3}{a}+\mathrm{2}{b}+{c}\mid\leqslant\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{3}{a}+\mathrm{2}{b}+{c}\in\left[−\mathrm{1},\mathrm{1}\right] \\ $$$$\Leftrightarrow\mathrm{3}{a}\in\left[−\left(\mathrm{2}{b}+{c}+\mathrm{1}\right),\mathrm{1}−\mathrm{2}{b}−{c}\right] \\ $$$$\Leftrightarrow{a}\in\left[−\left(\mathrm{2}{b}+{c}+\mathrm{1}\right)/\mathrm{3},\left(\mathrm{1}−\mathrm{2}{b}−{c}\right)/\mathrm{3}\right] \\ $$$$\mathrm{so}\:\mathrm{if}\:\mathrm{2}{b}+{c}<\mathrm{1}\:\mathrm{then}\:\mathrm{the}\:\mathrm{max} \\ $$$$\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{will}\:\mathrm{be}\:\:\frac{\mathrm{1}−\mathrm{2}{b}−{c}}{\mathrm{3}} \\ $$$$\left(\mathrm{if}\:\mathrm{2}{b}+{c}>\mathrm{1}\:\mathrm{we}\:\mathrm{just}\:\mathrm{don}'\mathrm{t}\:\mathrm{get}\:\mathrm{an}\:{a}\:\mathrm{from}\:\mathrm{this}\right. \\ $$$$\left.\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{formula}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{3}{a}+{b}\leqslant\mathrm{0}\:\Leftrightarrow{a}\in\left[\mathrm{0},−{b}/\mathrm{3}\right] \\ $$$$\left(\mathrm{so}\:{b}\:\mathrm{has}\:\mathrm{to}\:\mathrm{be}\:\mathrm{negtive}\right) \\ $$$$\mathrm{reusing}\:\mathrm{previous}\:\mathrm{computations}\:\mathrm{we}\:\mathrm{know}: \\ $$$${a}\in\left[\frac{−\mathrm{2}{b}−{c}−\mathrm{1}}{\mathrm{3}},\frac{−\mathrm{2}{b}−{c}+\mathrm{1}}{\mathrm{3}}\right] \\ $$$$\mathrm{if}\:\mathrm{2}{b}+{c}<\mathrm{1}\:\mathrm{we}\:\mathrm{get}\:\mathrm{that}\:\mathrm{the}\:\mathrm{max}\:\mathrm{value}\:\mathrm{of} \\ $$$${a}\:\mathrm{is}\:\mathrm{Min}\left\{−{b}/\mathrm{3},\:\left(−\mathrm{2}{b}−{c}+\mathrm{1}\right)/\mathrm{3}\right\} \\ $$$$=\frac{−\mathrm{2}{b}}{\mathrm{3}}+\mathrm{Min}\left\{{b}/\mathrm{3},\left(\mathrm{1}−{c}\right)/\mathrm{3}\right\} \\ $$$$\mathrm{Since}\:\mid{c}\mid\leqslant\mathrm{1}\:\mathrm{in}\:\mathrm{every}\:\mathrm{case},\:\mathrm{and}\:\mathrm{in}\:\mathrm{the}\:\mathrm{current} \\ $$$$\mathrm{case}\:{b}\leqslant\mathrm{0},\:\mathrm{it}\:\mathrm{is}\:\mathrm{obvious}\:\mathrm{that}\:\mathrm{we}\:\mathrm{get}: \\ $$$$\frac{\mathrm{1}−\mathrm{2}{b}−{c}}{\mathrm{3}}\:\mathrm{again}. \\ $$$$\mathrm{Last}\:\mathrm{but}\:\mathrm{not}\:\mathrm{least}\:{b}^{\mathrm{2}} \leqslant\mathrm{18}{a}\:\mathrm{does}\:\mathrm{not}\:\mathrm{constraint} \\ $$$$\mathrm{the}\:\mathrm{maximal}\:\mathrm{value}\:\mathrm{of}\:{a}.\:\mathrm{So}\:\mathrm{if}\:\mathrm{it}\:\mathrm{exist}\:\mathrm{in}\:\mathrm{the}\:\mathrm{case}\:\mathrm{of} \\ $$$$\mathrm{this}\:\mathrm{inequality},\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{also}\:\mathrm{has}\:\mathrm{to}\:\mathrm{be} \\ $$$$\frac{\mathrm{1}−\mathrm{2}{b}−{c}}{\mathrm{3}}\:\mathrm{again}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{To}\:\mathrm{conclude},\:\mathrm{in}\:\mathrm{the}\:\mathrm{case}\:\mathrm{of}\:\mathrm{the}\:\mathrm{requirement} \\ $$$${a}>\mathrm{0}\:\mathrm{the}\:\mathrm{maximal}\:\mathrm{value}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{taken}\:\mathrm{by} \\ $$$${a}\:\mathrm{while}\:\mid{f}\:'\left({x}\right)\mid\leqslant\mathrm{1}\:\mathrm{on}\:\left[\mathrm{0},\mathrm{1}\right]\:\mathrm{is}: \\ $$$${a}_{\mathrm{Max}} :=\:\frac{\mathrm{1}−\mathrm{2}{b}−{c}}{\mathrm{3}} \\ $$$$\mathrm{of}\:\mathrm{course},\:\mathrm{if}\:\mathrm{2}{b}+{c}\geqslant\mathrm{1}\:\mathrm{that}\:\mathrm{will}\:\mathrm{break}\:{a}>\mathrm{0}. \\ $$$$\mathrm{This}\:\mathrm{conclude}\:\mathrm{the}\:\mathrm{discussion}.\:_{\Box} \\ $$
Answered by Frix last updated on 10/Apr/24
![f′(x)=g(x)=3ax^2 +2bx+c a → max ⇒ a>0 ⇒ g(x) has an absolute min at x_0 =−(b/(3a)) Case 1: −(b/(3a))∉]0, 1[ To get the steepest possible parabola let g(0)=−1∧g(1)=1 ( _(the same a)^(g(0)=1∧g(1)=−1 leads to) ) ⇒ b=1−((3a)/2)∧c=−1 ⇒ x_0 =(1/2)−(1/(3a)) x_0 ≤0∨1≤x_0 ⇒ a≤(2/3) witb a=(2/3) g(x)=f′(x)=2x^2 −1 (or 2x^2 −4x+1) Case 2: −(b/(3a))∈]0, 1[ a>0⇒b<0 g(−(b/(3a)))=−1∧g(1)=1 () ⇒ b=(√(6a))−3a∧c=3a−2(√(6a))+1 ⇒ x_0 =1−((√2)/( (√(3a)))) 0<x_0 <1 ⇒ a>(2/3) g(1)=1 ⇒ −1≤g(0)≤1 g(0)=c ⇒ −1≤3a−2(√(6a))+1≤1 ⇒ 0≤a≤(8/3) ⇒ a_(max) =(8/3) g(x)=f′(x)=8x^2 −8x+1](https://www.tinkutara.com/question/Q206274.png)
$${f}'\left({x}\right)={g}\left({x}\right)=\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{2}{bx}+{c} \\ $$$${a}\:\rightarrow\:\mathrm{max}\:\Rightarrow\:{a}>\mathrm{0}\:\Rightarrow \\ $$$${g}\left({x}\right)\:\mathrm{has}\:\mathrm{an}\:\mathrm{absolute}\:\mathrm{min}\:\mathrm{at}\:{x}_{\mathrm{0}} =−\frac{{b}}{\mathrm{3}{a}} \\ $$$$ \\ $$$$\left.\mathrm{Case}\:\mathrm{1}:\:−\frac{{b}}{\mathrm{3}{a}}\notin\right]\mathrm{0},\:\mathrm{1}\left[\right. \\ $$$$\mathrm{To}\:\mathrm{get}\:\mathrm{the}\:\mathrm{steepest}\:\mathrm{possible}\:\mathrm{parabola}\:\mathrm{let} \\ $$$${g}\left(\mathrm{0}\right)=−\mathrm{1}\wedge{g}\left(\mathrm{1}\right)=\mathrm{1}\:\left(\:_{\mathrm{the}\:\mathrm{same}\:{a}} ^{{g}\left(\mathrm{0}\right)=\mathrm{1}\wedge{g}\left(\mathrm{1}\right)=−\mathrm{1}\:\mathrm{leads}\:\mathrm{to}} \right) \\ $$$$\Rightarrow \\ $$$${b}=\mathrm{1}−\frac{\mathrm{3}{a}}{\mathrm{2}}\wedge{c}=−\mathrm{1}\:\Rightarrow\:{x}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}{a}} \\ $$$${x}_{\mathrm{0}} \leqslant\mathrm{0}\vee\mathrm{1}\leqslant{x}_{\mathrm{0}} \:\Rightarrow\:{a}\leqslant\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{witb}\:{a}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${g}\left({x}\right)={f}'\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\:\left(\mathrm{or}\:\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}\right) \\ $$$$ \\ $$$$\left.\mathrm{Case}\:\mathrm{2}:\:−\frac{{b}}{\mathrm{3}{a}}\in\right]\mathrm{0},\:\mathrm{1}\left[\right. \\ $$$${a}>\mathrm{0}\Rightarrow{b}<\mathrm{0} \\ $$$${g}\left(−\frac{{b}}{\mathrm{3}{a}}\right)=−\mathrm{1}\wedge{g}\left(\mathrm{1}\right)=\mathrm{1}\:\left(\right) \\ $$$$\Rightarrow \\ $$$${b}=\sqrt{\mathrm{6}{a}}−\mathrm{3}{a}\wedge{c}=\mathrm{3}{a}−\mathrm{2}\sqrt{\mathrm{6}{a}}+\mathrm{1}\:\Rightarrow\:{x}_{\mathrm{0}} =\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}{a}}} \\ $$$$\mathrm{0}<{x}_{\mathrm{0}} <\mathrm{1}\:\Rightarrow\:{a}>\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${g}\left(\mathrm{1}\right)=\mathrm{1}\:\Rightarrow\:−\mathrm{1}\leqslant{g}\left(\mathrm{0}\right)\leqslant\mathrm{1} \\ $$$${g}\left(\mathrm{0}\right)={c}\:\Rightarrow\:−\mathrm{1}\leqslant\mathrm{3}{a}−\mathrm{2}\sqrt{\mathrm{6}{a}}+\mathrm{1}\leqslant\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{0}\leqslant{a}\leqslant\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\Rightarrow\:{a}_{\mathrm{max}} =\frac{\mathrm{8}}{\mathrm{3}} \\ $$$${g}\left({x}\right)={f}'\left({x}\right)=\mathrm{8}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{1} \\ $$