Showing that a function is increasing and then proving an inequality












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Let $f:[0, infty) to mathbb{R} $, $f(x) =3^x+10^x+21^x-6^x-7^x-15^x$. Prove that :

a) $f$ is an increasing function.

b) $2cdot 3^x+9cdot 10^x+20cdot 21^xge 5cdot 6^x+6cdot 7^x+14cdot 15^x$,$forall xge 0$

I tried differentiating the function, but it didn't help. Then I thought about factoring the function, but this didn't work either.






real-analysis algebra-precalculus inequality







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    0












    $begingroup$


    Let $f:[0, infty) to mathbb{R} $, $f(x) =3^x+10^x+21^x-6^x-7^x-15^x$. Prove that :

    a) $f$ is an increasing function.

    b) $2cdot 3^x+9cdot 10^x+20cdot 21^xge 5cdot 6^x+6cdot 7^x+14cdot 15^x$,$forall xge 0$

    I tried differentiating the function, but it didn't help. Then I thought about factoring the function, but this didn't work either.






    real-analysis algebra-precalculus inequality







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      0



      $begingroup$


      Let $f:[0, infty) to mathbb{R} $, $f(x) =3^x+10^x+21^x-6^x-7^x-15^x$. Prove that :

      a) $f$ is an increasing function.

      b) $2cdot 3^x+9cdot 10^x+20cdot 21^xge 5cdot 6^x+6cdot 7^x+14cdot 15^x$,$forall xge 0$

      I tried differentiating the function, but it didn't help. Then I thought about factoring the function, but this didn't work either.






      real-analysis algebra-precalculus inequality







      share|cite|improve this question











      $endgroup$




      Let $f:[0, infty) to mathbb{R} $, $f(x) =3^x+10^x+21^x-6^x-7^x-15^x$. Prove that :

      a) $f$ is an increasing function.

      b) $2cdot 3^x+9cdot 10^x+20cdot 21^xge 5cdot 6^x+6cdot 7^x+14cdot 15^x$,$forall xge 0$

      I tried differentiating the function, but it didn't help. Then I thought about factoring the function, but this didn't work either.






      real-analysis algebra-precalculus inequality




      real-analysis algebra-precalculus inequality






      share|cite|improve this question















      share|cite|improve this question













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      edited Feb 2 at 18:41







      MathEnthusiast

















      asked Feb 2 at 18:26









      MathEnthusiastMathEnthusiast

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          $begingroup$

          We can write
          $$
          f(x) = (3^x-2^x)(7^x-5^x)+(2^x-1)(7^x-3^x).
          $$           It can be easily seen that each factor is a non-negative, increasing function, for example,
          $$
          3^x-2^x = 2^xleft( left (frac{3}{2}right)^x-1right)
          $$           is a product of two non-negative, increasing functions. Hence $f$ is also an incresing function.



          The second assertion is equivalent to $f(x+1)-f(x)ge 0$. This is true since $f$ is increasing.






          share|cite|improve this answer









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            $begingroup$

            We can write
            $$
            f(x) = (3^x-2^x)(7^x-5^x)+(2^x-1)(7^x-3^x).
            $$             It can be easily seen that each factor is a non-negative, increasing function, for example,
            $$
            3^x-2^x = 2^xleft( left (frac{3}{2}right)^x-1right)
            $$             is a product of two non-negative, increasing functions. Hence $f$ is also an incresing function.



            The second assertion is equivalent to $f(x+1)-f(x)ge 0$. This is true since $f$ is increasing.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              We can write
              $$
              f(x) = (3^x-2^x)(7^x-5^x)+(2^x-1)(7^x-3^x).
              $$               It can be easily seen that each factor is a non-negative, increasing function, for example,
              $$
              3^x-2^x = 2^xleft( left (frac{3}{2}right)^x-1right)
              $$               is a product of two non-negative, increasing functions. Hence $f$ is also an incresing function.



              The second assertion is equivalent to $f(x+1)-f(x)ge 0$. This is true since $f$ is increasing.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                We can write
                $$
                f(x) = (3^x-2^x)(7^x-5^x)+(2^x-1)(7^x-3^x).
                $$                 It can be easily seen that each factor is a non-negative, increasing function, for example,
                $$
                3^x-2^x = 2^xleft( left (frac{3}{2}right)^x-1right)
                $$                 is a product of two non-negative, increasing functions. Hence $f$ is also an incresing function.



                The second assertion is equivalent to $f(x+1)-f(x)ge 0$. This is true since $f$ is increasing.






                share|cite|improve this answer









                $endgroup$



                We can write
                $$
                f(x) = (3^x-2^x)(7^x-5^x)+(2^x-1)(7^x-3^x).
                $$                 It can be easily seen that each factor is a non-negative, increasing function, for example,
                $$
                3^x-2^x = 2^xleft( left (frac{3}{2}right)^x-1right)
                $$                 is a product of two non-negative, increasing functions. Hence $f$ is also an incresing function.



                The second assertion is equivalent to $f(x+1)-f(x)ge 0$. This is true since $f$ is increasing.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 2 at 20:56









                SongSong

                18.6k21651




                18.6k21651






























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