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AlgebraQuestion and Answers: Page 266

Question Number 80209    Answers: 2   Comments: 1

Question Number 80208    Answers: 0   Comments: 0

Question Number 80204    Answers: 0   Comments: 2

Question Number 80199    Answers: 2   Comments: 1

Question Number 80178    Answers: 0   Comments: 4

Question Number 80161    Answers: 0   Comments: 10

Find the relation between q and r so that x^3 +3px^2 +qx+r is a perfect cube for all value of x

$${Find}\:{the}\:{relation}\:{between} \\ $$$${q}\:{and}\:{r}\:\:{so}\:\:{that} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}\:{is}\:{a}\:{perfect} \\ $$$${cube}\:{for}\:{all}\:\:{value}\:{of}\:{x} \\ $$

Question Number 80160    Answers: 0   Comments: 1

∫_0 ^∞ (x^3 /(e^(2x) −e^x ))

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{{e}^{\mathrm{2}{x}} −{e}^{{x}} } \\ $$

Question Number 80139    Answers: 1   Comments: 6

Question Number 80142    Answers: 2   Comments: 0

a. Σ_(k=1) ^∞ ((k^3 /2^k ))=? b. Σ_(k=1) ^∞ (((k^3 +k^2 +k+1)/7^k ))=?

$$\mathrm{a}.\:\:\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\left(\frac{\boldsymbol{\mathrm{k}}^{\mathrm{3}} }{\mathrm{2}^{\boldsymbol{\mathrm{k}}} }\right)=? \\ $$$$\boldsymbol{\mathrm{b}}.\:\:\:\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\left(\frac{\boldsymbol{\mathrm{k}}^{\mathrm{3}} +\boldsymbol{\mathrm{k}}^{\mathrm{2}} +\boldsymbol{\mathrm{k}}+\mathrm{1}}{\mathrm{7}^{\boldsymbol{\mathrm{k}}} }\right)=? \\ $$

Question Number 80131    Answers: 0   Comments: 2

Question Number 80145    Answers: 1   Comments: 5

{ (((x/a)+(y/b)=a^2 +b^2 )),(( [a,b∈R])),((ab(x^2 −y^2 )=xy(a^2 −b^2 ))) :}

$$\begin{cases}{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{a}}}+\frac{\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{b}}}=\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} }\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\boldsymbol{\mathrm{R}}\right]}\\{\boldsymbol{\mathrm{ab}}\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)=\boldsymbol{\mathrm{xy}}\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} \right)}\end{cases} \\ $$

Question Number 80144    Answers: 0   Comments: 1

solve for x: (((√x)+1)/(√(x+1)))+ax^2 =x(a^2 +1) [a∈R]

$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}}: \\ $$$$\frac{\sqrt{\boldsymbol{\mathrm{x}}}+\mathrm{1}}{\sqrt{\boldsymbol{\mathrm{x}}+\mathrm{1}}}+\boldsymbol{\mathrm{ax}}^{\mathrm{2}} =\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{1}\right)\:\:\:\:\:\:\left[\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}\right] \\ $$

Question Number 80116    Answers: 1   Comments: 1

Find S_m =Σ_(n=0) ^∞ (1/(Π_(k=1) ^m (n+k)))=? (m≥2)

$${Find} \\ $$$${S}_{{m}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\left({n}+{k}\right)}=? \\ $$$$\left({m}\geqslant\mathrm{2}\right) \\ $$

Question Number 80093    Answers: 3   Comments: 0

Solve for x and y x^(√y) = 64 y^(√x) = 81

$$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$$$\:\:\:\:\:\mathrm{x}^{\sqrt{\mathrm{y}}} \:\:\:=\:\:\mathrm{64} \\ $$$$\:\:\:\:\:\mathrm{y}^{\sqrt{\mathrm{x}}} \:\:\:=\:\mathrm{81} \\ $$

Question Number 80084    Answers: 0   Comments: 3

−1=(−1)^1 =(−1)^(2/2) =((−1)^2 )^(1/2) =(1)^(1/2) = =(√1)=1 what do you think about this?

$$\:\:−\mathrm{1}=\left(−\mathrm{1}\right)^{\mathrm{1}} =\left(−\mathrm{1}\right)^{\frac{\mathrm{2}}{\mathrm{2}}} =\left(\left(−\mathrm{1}\right)^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\left(\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} = \\ $$$$=\sqrt{\mathrm{1}}=\mathrm{1}\:\: \\ $$$$\mathrm{what}\:\mathrm{do}\:\mathrm{you}\:\mathrm{think}\:\mathrm{about}\:\mathrm{this}? \\ $$

Question Number 80068    Answers: 2   Comments: 3

Question Number 80053    Answers: 0   Comments: 4

Find integer x, y such that 2^x −y^2 =615

$${Find}\:{integer}\:{x},\:{y}\:{such}\:{that} \\ $$$$\mathrm{2}^{{x}} −{y}^{\mathrm{2}} =\mathrm{615} \\ $$

Question Number 80108    Answers: 1   Comments: 3

a,b,c ∈R ((b+c+d)/a)=((a+c+d)/b)=((a+b+c)/d)=((a+b+d)/c)=r what is r?

$${a},{b},{c}\:\in\mathbb{R} \\ $$$$\frac{{b}+{c}+{d}}{{a}}=\frac{{a}+{c}+{d}}{{b}}=\frac{{a}+{b}+{c}}{{d}}=\frac{{a}+{b}+{d}}{{c}}={r} \\ $$$${what}\:{is}\:{r}? \\ $$

Question Number 80039    Answers: 1   Comments: 6

prove that (1+x)(1+(1/x))≥4

$${prove}\:{that} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\geqslant\mathrm{4} \\ $$

Question Number 79998    Answers: 0   Comments: 3

given 3x + 4y+1 = 3(√x) + 2(√y) find the value of (√(x.y))

$$\mathrm{given}\:\mathrm{3x}\:+\:\mathrm{4y}+\mathrm{1}\:=\:\mathrm{3}\sqrt{\mathrm{x}}\:+\:\mathrm{2}\sqrt{\mathrm{y}}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\sqrt{\mathrm{x}.\mathrm{y}}\: \\ $$

Question Number 79978    Answers: 3   Comments: 0

Given for x,y,z>0: 2^x =3^y =5^z Arrange 2x, 3y, 5z in increasing order.

$${Given}\:{for}\:{x},{y},{z}>\mathrm{0}: \\ $$$$\mathrm{2}^{{x}} =\mathrm{3}^{{y}} =\mathrm{5}^{{z}} \\ $$$${Arrange}\:\mathrm{2}{x},\:\mathrm{3}{y},\:\mathrm{5}{z}\:{in}\:{increasing}\:{order}. \\ $$

Question Number 79974    Answers: 1   Comments: 0

Question Number 80004    Answers: 1   Comments: 2

x and y any integer satisfy equation (x−2004)(x−2006)=2^y the greatest possible value of x+y

$${x}\:{and}\:{y}\:{any}\:{integer}\:{satisfy} \\ $$$${equation}\:\left({x}−\mathrm{2004}\right)\left({x}−\mathrm{2006}\right)=\mathrm{2}^{{y}} \\ $$$${the}\:{greatest}\:{possible}\:{value} \\ $$$${of}\:{x}+{y} \\ $$

Question Number 79932    Answers: 1   Comments: 1

Question Number 79883    Answers: 0   Comments: 9

to Sir Jagoll (and of course everybody else) (1) y=((x^2 −x−6)/(x^2 −3x−4))= =(((x−3)(x+2))/((x−4)(x+1))) ⇒ ⇒ { ((zeros at x=−2; x=3)),((vertical asymptotes at x=−1; x=4)) :} defined for x∈R\{−1; 4} range: transforming y=((x^2 −x−6)/(x^2 −3x−4)) to x^2 −((3y−1)/(y−1))x−((2(2y−3))/(y−1))=0 D=((25y^2 −46y+25)/(4(y−1)^2 ))>0∀x∈R ⇒ ⇒ { ((range=R)),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=1 ⇒ y=1 y′=−((2(x^2 −2x+7))/((x−4)^2 (x+1)^2 )) no real zeros ⇒ ⇒ no local extremes y′′=((4(x^3 −3x^2 +21x−25))/((x−4)^3 (x+1)^3 ))=0 at x≈1.33131 ⇒ ⇒ turning point (2) y=((x^2 −x−6)/(x^2 −3x+4))= =(((x−3)(x+2))/(x^2 −3x+4)); x^2 −3x+4=0 no real zeros ⇒ ⇒ { ((zeros at x=−2; x=3)),((no vertical asymptote)) :} defined for x∈R range: transforming y=((x^2 −x−6)/(x^2 −3x+4)) to x^2 −((3y−1)/(y−1))x+((2(2y+3))/(y−1))=0 D=−((7y^2 +14y−25)/(4(y−1)^2 ))≥0 for −1−((4(√(14)))/7)≤y≤−1+((4(√(14)))/7) ⇒ ⇒ { ((range=[−1−((4(√(14)))/7); −1+((4(√(14)))/7)])),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=5 ⇒ y=1 y′=−((2(x^2 −10x+11))/((x^2 −3x+4)^2 ))=0 at x=5±(√(14)) y^(′′) =((4(x^3 −15x^2 +33x−13))/((x^2 −3x+4)^3 )) y′′ { ((>0 at x=5−(√(14)) ⇒ local minimum)),((<0 at x=5+(√(14)) ⇒ local maximum)),((=0 at { ((x≈.506699)),((x≈2.06421)),((x≈12.4291)) :} ⇒ 3 turning points )) :} (3) y=((x^2 −x+6)/(x^2 −3x−4))= =((x^2 −x+6)/((x−4)(x+1))) ⇒ ⇒ { ((no real zeros)),((vertical asymptotes at x=−1; x=4)) :} defined for x∈R\{−1; 4} range: transforming y=((x^2 −x+6)/(x^2 −3x−4)) to x^2 −((3y−1)/(y−1))x−((2(2y+3))/(y−1))=0 D=((25y^2 +2y−23)/(4(y−1)^2 ))<0 for −1<y<((23)/(25)) ⇒ ⇒ { ((range=R\]−1; ((23)/(25))[)),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=−5 ⇒ y=1 y′=−((2(x^2 +10x−11))/((x−4)^2 (x+1)^2 ))=0 at x=−11; x=1 y′′=((4(x^3 +15x^2 −33x+53))/((x−4)^3 (x+1)^3 )) y′′ { ((>0 at x=−11 ⇒ local minimum)),((<0 at x=1 ⇒ local maximum)),((=0 at x≈−17.1098 ⇒ turning point)) :}

$$\mathrm{to}\:\mathrm{Sir}\:\mathrm{Jagoll}\:\left({and}\:{of}\:{course}\:{everybody}\:{else}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}= \\ $$$$=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{zeros}\:\mathrm{at}\:{x}=−\mathrm{2};\:{x}=\mathrm{3}}\\{\mathrm{vertical}\:\mathrm{asymptotes}\:\mathrm{at}\:{x}=−\mathrm{1};\:{x}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1};\:\mathrm{4}\right\} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}−\frac{\mathrm{2}\left(\mathrm{2}{y}−\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{25}{y}^{\mathrm{2}} −\mathrm{46}{y}+\mathrm{25}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }>\mathrm{0}\forall{x}\in\mathbb{R}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{range}=\mathbb{R}}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{1}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{7}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{local}\:\mathrm{extremes} \\ $$$${y}''=\frac{\mathrm{4}\left({x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{21}{x}−\mathrm{25}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{0}\:\mathrm{at}\:{x}\approx\mathrm{1}.\mathrm{33131}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{turning}\:\mathrm{point} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}}= \\ $$$$=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}};\:{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}=\mathrm{0}\:\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{zeros}\:\mathrm{at}\:{x}=−\mathrm{2};\:{x}=\mathrm{3}}\\{\mathrm{no}\:\mathrm{vertical}\:\mathrm{asymptote}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}+\frac{\mathrm{2}\left(\mathrm{2}{y}+\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=−\frac{\mathrm{7}{y}^{\mathrm{2}} +\mathrm{14}{y}−\mathrm{25}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }\geqslant\mathrm{0}\:\mathrm{for}\:−\mathrm{1}−\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\leqslant{y}\leqslant−\mathrm{1}+\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{range}=\left[−\mathrm{1}−\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}};\:−\mathrm{1}+\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\right]}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{5}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{11}\right)}{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}\pm\sqrt{\mathrm{14}} \\ $$$${y}^{''} =\frac{\mathrm{4}\left({x}^{\mathrm{3}} −\mathrm{15}{x}^{\mathrm{2}} +\mathrm{33}{x}−\mathrm{13}\right)}{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{3}} } \\ $$$${y}''\begin{cases}{>\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}−\sqrt{\mathrm{14}}\:\Rightarrow\:\mathrm{local}\:\mathrm{minimum}}\\{<\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}+\sqrt{\mathrm{14}}\:\Rightarrow\:\mathrm{local}\:\mathrm{maximum}}\\{=\mathrm{0}\:\mathrm{at}\:\begin{cases}{{x}\approx.\mathrm{506699}}\\{{x}\approx\mathrm{2}.\mathrm{06421}}\\{{x}\approx\mathrm{12}.\mathrm{4291}}\end{cases}\:\Rightarrow\:\mathrm{3}\:\mathrm{turning}\:\mathrm{points}\:}\end{cases} \\ $$$$ \\ $$$$\left(\mathrm{3}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}= \\ $$$$=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}}\\{\mathrm{vertical}\:\mathrm{asymptotes}\:\mathrm{at}\:{x}=−\mathrm{1};\:{x}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1};\:\mathrm{4}\right\} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}−\frac{\mathrm{2}\left(\mathrm{2}{y}+\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{25}{y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{23}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }<\mathrm{0}\:\mathrm{for}\:−\mathrm{1}<{y}<\frac{\mathrm{23}}{\mathrm{25}}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\left.\mathrm{range}=\mathbb{R}\backslash\right]−\mathrm{1};\:\frac{\mathrm{23}}{\mathrm{25}}\left[\right.}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=−\mathrm{5}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{11}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0}\:\mathrm{at}\:{x}=−\mathrm{11};\:{x}=\mathrm{1} \\ $$$${y}''=\frac{\mathrm{4}\left({x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} −\mathrm{33}{x}+\mathrm{53}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${y}''\begin{cases}{>\mathrm{0}\:\mathrm{at}\:{x}=−\mathrm{11}\:\Rightarrow\:\mathrm{local}\:\mathrm{minimum}}\\{<\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{1}\:\Rightarrow\:\mathrm{local}\:\mathrm{maximum}}\\{=\mathrm{0}\:\mathrm{at}\:{x}\approx−\mathrm{17}.\mathrm{1098}\:\Rightarrow\:\mathrm{turning}\:\mathrm{point}}\end{cases} \\ $$

Question Number 79876    Answers: 1   Comments: 1

f is derivable in R. 1) Demonstrate that if f is pair , f ′ is odd(unpair). 1) Demonstrate that if f is unpair , f ′ is pair. Please help me sirs

$${f}\:{is}\:{derivable}\:{in}\:\mathbb{R}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Demonstrate}\:\mathrm{that}\:\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{pair}\:,\:{f}\:'\:\mathrm{is}\:\mathrm{odd}\left(\mathrm{unpair}\right). \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Demonstrate}\:\mathrm{that}\:\mathrm{if}\:{f}\:{is}\:\mathrm{unpair}\:,\:{f}\:'\:\mathrm{is}\:\mathrm{pair}. \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sirs} \\ $$

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