Find the value of $a$











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Find $a$ for which $f(x) = left(frac{sqrt{a+4}}{1-a} -1right)x^5-3x+ln5 ;$ decreases for all $x$ with $aneq 1$ and $ageq -4$.




My try:



For $f(x)$ to decrease
$$5x^4left(frac{sqrt{a+4}}{1-a} -1right) -3 <0implies x^4left(frac{sqrt{a+4}}{1-a} -1right) < 3/5 $$



How can I proceed further?






calculus functions







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    up vote
    1
    down vote

    favorite













    Find $a$ for which $f(x) = left(frac{sqrt{a+4}}{1-a} -1right)x^5-3x+ln5 ;$ decreases for all $x$ with $aneq 1$ and $ageq -4$.




    My try:



    For $f(x)$ to decrease
    $$5x^4left(frac{sqrt{a+4}}{1-a} -1right) -3 <0implies x^4left(frac{sqrt{a+4}}{1-a} -1right) < 3/5 $$



    How can I proceed further?






    calculus functions







    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite












      Find $a$ for which $f(x) = left(frac{sqrt{a+4}}{1-a} -1right)x^5-3x+ln5 ;$ decreases for all $x$ with $aneq 1$ and $ageq -4$.




      My try:



      For $f(x)$ to decrease
      $$5x^4left(frac{sqrt{a+4}}{1-a} -1right) -3 <0implies x^4left(frac{sqrt{a+4}}{1-a} -1right) < 3/5 $$



      How can I proceed further?






      calculus functions







      share|cite|improve this question
















      Find $a$ for which $f(x) = left(frac{sqrt{a+4}}{1-a} -1right)x^5-3x+ln5 ;$ decreases for all $x$ with $aneq 1$ and $ageq -4$.




      My try:



      For $f(x)$ to decrease
      $$5x^4left(frac{sqrt{a+4}}{1-a} -1right) -3 <0implies x^4left(frac{sqrt{a+4}}{1-a} -1right) < 3/5 $$



      How can I proceed further?






      calculus functions




      calculus functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 23 at 20:44









      Yadati Kiran

      1,499518




      1,499518










      asked Nov 23 at 20:27









      Md Masood

      427




      427






















          2 Answers
          2






          active

          oldest

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          up vote
          1
          down vote



          accepted










          This holds true for every $x$ only if $$frac{sqrt{a+4}}{1-a} -1le 0$$or $${sqrt{a+4}over 1-a}le 1$$Surely $a>1$ is one region of answer. For $a<1$ we have $$sqrt {a+4}le 1-ato a+4le a^2-2a+1to a^2-3a-3ge 0to\age {3+sqrt {21}over 2}\ale {3-sqrt {21}over 2}$$therefore the region of the answer is $$[-4, {3-sqrt {21}over 2}]cup (1,infty)$$






          share|cite|improve this answer





















          • Why the coefficent of $x^4$ need to be negative.
            – Md Masood
            Nov 23 at 21:17








          • 1




            If it is positive then for large enough $x$ the inequality doesn't hold.
            – Mostafa Ayaz
            Nov 23 at 21:18










          • Oh yes , thanks.
            – Md Masood
            Nov 23 at 21:21






          • 1




            You're welcome.......
            – Mostafa Ayaz
            Nov 23 at 21:22


















          up vote
          2
          down vote













          HINT



          We have



          $$f(x) = left(frac{sqrt{a+4}}{1-a} -1right)x^5-3x+ln5 implies f'(x) = 5left(frac{sqrt{a+4}}{1-a} -1right)x^4-3le 0$$



          then we need



          $$left(frac{sqrt{a+4}}{1-a} -1right)x^4- frac35 le 0 $$



          that is true for all $x$ when



          $$frac{sqrt{a+4}}{1-a} -1le0 iff frac{sqrt{a+4}+a-1}{1-a}le0$$






          share|cite|improve this answer























          • oh no, @gimusi, fix it...
            – user376343
            Nov 23 at 20:43










          • Ummm... $f(x)$ decreases...
            – David G. Stork
            Nov 23 at 20:44










          • Opsss...I was dealing with the incresing case! I fix, thanks
            – gimusi
            Nov 23 at 20:44











          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          up vote
          1
          down vote



          accepted










          This holds true for every $x$ only if $$frac{sqrt{a+4}}{1-a} -1le 0$$or $${sqrt{a+4}over 1-a}le 1$$Surely $a>1$ is one region of answer. For $a<1$ we have $$sqrt {a+4}le 1-ato a+4le a^2-2a+1to a^2-3a-3ge 0to\age {3+sqrt {21}over 2}\ale {3-sqrt {21}over 2}$$therefore the region of the answer is $$[-4, {3-sqrt {21}over 2}]cup (1,infty)$$






          share|cite|improve this answer





















          • Why the coefficent of $x^4$ need to be negative.
            – Md Masood
            Nov 23 at 21:17








          • 1




            If it is positive then for large enough $x$ the inequality doesn't hold.
            – Mostafa Ayaz
            Nov 23 at 21:18










          • Oh yes , thanks.
            – Md Masood
            Nov 23 at 21:21






          • 1




            You're welcome.......
            – Mostafa Ayaz
            Nov 23 at 21:22















          up vote
          1
          down vote



          accepted










          This holds true for every $x$ only if $$frac{sqrt{a+4}}{1-a} -1le 0$$or $${sqrt{a+4}over 1-a}le 1$$Surely $a>1$ is one region of answer. For $a<1$ we have $$sqrt {a+4}le 1-ato a+4le a^2-2a+1to a^2-3a-3ge 0to\age {3+sqrt {21}over 2}\ale {3-sqrt {21}over 2}$$therefore the region of the answer is $$[-4, {3-sqrt {21}over 2}]cup (1,infty)$$






          share|cite|improve this answer





















          • Why the coefficent of $x^4$ need to be negative.
            – Md Masood
            Nov 23 at 21:17








          • 1




            If it is positive then for large enough $x$ the inequality doesn't hold.
            – Mostafa Ayaz
            Nov 23 at 21:18










          • Oh yes , thanks.
            – Md Masood
            Nov 23 at 21:21






          • 1




            You're welcome.......
            – Mostafa Ayaz
            Nov 23 at 21:22













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          This holds true for every $x$ only if $$frac{sqrt{a+4}}{1-a} -1le 0$$or $${sqrt{a+4}over 1-a}le 1$$Surely $a>1$ is one region of answer. For $a<1$ we have $$sqrt {a+4}le 1-ato a+4le a^2-2a+1to a^2-3a-3ge 0to\age {3+sqrt {21}over 2}\ale {3-sqrt {21}over 2}$$therefore the region of the answer is $$[-4, {3-sqrt {21}over 2}]cup (1,infty)$$






          share|cite|improve this answer












          This holds true for every $x$ only if $$frac{sqrt{a+4}}{1-a} -1le 0$$or $${sqrt{a+4}over 1-a}le 1$$Surely $a>1$ is one region of answer. For $a<1$ we have $$sqrt {a+4}le 1-ato a+4le a^2-2a+1to a^2-3a-3ge 0to\age {3+sqrt {21}over 2}\ale {3-sqrt {21}over 2}$$therefore the region of the answer is $$[-4, {3-sqrt {21}over 2}]cup (1,infty)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 at 20:54









          Mostafa Ayaz

          13.6k3836




          13.6k3836












          • Why the coefficent of $x^4$ need to be negative.
            – Md Masood
            Nov 23 at 21:17








          • 1




            If it is positive then for large enough $x$ the inequality doesn't hold.
            – Mostafa Ayaz
            Nov 23 at 21:18










          • Oh yes , thanks.
            – Md Masood
            Nov 23 at 21:21






          • 1




            You're welcome.......
            – Mostafa Ayaz
            Nov 23 at 21:22


















          • Why the coefficent of $x^4$ need to be negative.
            – Md Masood
            Nov 23 at 21:17








          • 1




            If it is positive then for large enough $x$ the inequality doesn't hold.
            – Mostafa Ayaz
            Nov 23 at 21:18










          • Oh yes , thanks.
            – Md Masood
            Nov 23 at 21:21






          • 1




            You're welcome.......
            – Mostafa Ayaz
            Nov 23 at 21:22
















          Why the coefficent of $x^4$ need to be negative.
          – Md Masood
          Nov 23 at 21:17






          Why the coefficent of $x^4$ need to be negative.
          – Md Masood
          Nov 23 at 21:17






          1




          1




          If it is positive then for large enough $x$ the inequality doesn't hold.
          – Mostafa Ayaz
          Nov 23 at 21:18




          If it is positive then for large enough $x$ the inequality doesn't hold.
          – Mostafa Ayaz
          Nov 23 at 21:18












          Oh yes , thanks.
          – Md Masood
          Nov 23 at 21:21




          Oh yes , thanks.
          – Md Masood
          Nov 23 at 21:21




          1




          1




          You're welcome.......
          – Mostafa Ayaz
          Nov 23 at 21:22




          You're welcome.......
          – Mostafa Ayaz
          Nov 23 at 21:22










          up vote
          2
          down vote













          HINT



          We have



          $$f(x) = left(frac{sqrt{a+4}}{1-a} -1right)x^5-3x+ln5 implies f'(x) = 5left(frac{sqrt{a+4}}{1-a} -1right)x^4-3le 0$$



          then we need



          $$left(frac{sqrt{a+4}}{1-a} -1right)x^4- frac35 le 0 $$



          that is true for all $x$ when



          $$frac{sqrt{a+4}}{1-a} -1le0 iff frac{sqrt{a+4}+a-1}{1-a}le0$$






          share|cite|improve this answer























          • oh no, @gimusi, fix it...
            – user376343
            Nov 23 at 20:43










          • Ummm... $f(x)$ decreases...
            – David G. Stork
            Nov 23 at 20:44










          • Opsss...I was dealing with the incresing case! I fix, thanks
            – gimusi
            Nov 23 at 20:44















          up vote
          2
          down vote













          HINT



          We have



          $$f(x) = left(frac{sqrt{a+4}}{1-a} -1right)x^5-3x+ln5 implies f'(x) = 5left(frac{sqrt{a+4}}{1-a} -1right)x^4-3le 0$$



          then we need



          $$left(frac{sqrt{a+4}}{1-a} -1right)x^4- frac35 le 0 $$



          that is true for all $x$ when



          $$frac{sqrt{a+4}}{1-a} -1le0 iff frac{sqrt{a+4}+a-1}{1-a}le0$$






          share|cite|improve this answer























          • oh no, @gimusi, fix it...
            – user376343
            Nov 23 at 20:43










          • Ummm... $f(x)$ decreases...
            – David G. Stork
            Nov 23 at 20:44










          • Opsss...I was dealing with the incresing case! I fix, thanks
            – gimusi
            Nov 23 at 20:44













          up vote
          2
          down vote










          up vote
          2
          down vote









          HINT



          We have



          $$f(x) = left(frac{sqrt{a+4}}{1-a} -1right)x^5-3x+ln5 implies f'(x) = 5left(frac{sqrt{a+4}}{1-a} -1right)x^4-3le 0$$



          then we need



          $$left(frac{sqrt{a+4}}{1-a} -1right)x^4- frac35 le 0 $$



          that is true for all $x$ when



          $$frac{sqrt{a+4}}{1-a} -1le0 iff frac{sqrt{a+4}+a-1}{1-a}le0$$






          share|cite|improve this answer














          HINT



          We have



          $$f(x) = left(frac{sqrt{a+4}}{1-a} -1right)x^5-3x+ln5 implies f'(x) = 5left(frac{sqrt{a+4}}{1-a} -1right)x^4-3le 0$$



          then we need



          $$left(frac{sqrt{a+4}}{1-a} -1right)x^4- frac35 le 0 $$



          that is true for all $x$ when



          $$frac{sqrt{a+4}}{1-a} -1le0 iff frac{sqrt{a+4}+a-1}{1-a}le0$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 23 at 20:48

























          answered Nov 23 at 20:41









          gimusi

          1




          1












          • oh no, @gimusi, fix it...
            – user376343
            Nov 23 at 20:43










          • Ummm... $f(x)$ decreases...
            – David G. Stork
            Nov 23 at 20:44










          • Opsss...I was dealing with the incresing case! I fix, thanks
            – gimusi
            Nov 23 at 20:44


















          • oh no, @gimusi, fix it...
            – user376343
            Nov 23 at 20:43










          • Ummm... $f(x)$ decreases...
            – David G. Stork
            Nov 23 at 20:44










          • Opsss...I was dealing with the incresing case! I fix, thanks
            – gimusi
            Nov 23 at 20:44
















          oh no, @gimusi, fix it...
          – user376343
          Nov 23 at 20:43




          oh no, @gimusi, fix it...
          – user376343
          Nov 23 at 20:43












          Ummm... $f(x)$ decreases...
          – David G. Stork
          Nov 23 at 20:44




          Ummm... $f(x)$ decreases...
          – David G. Stork
          Nov 23 at 20:44












          Opsss...I was dealing with the incresing case! I fix, thanks
          – gimusi
          Nov 23 at 20:44




          Opsss...I was dealing with the incresing case! I fix, thanks
          – gimusi
          Nov 23 at 20:44


















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