how to find the interval at which a derivative function is increasing












0












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Alright, so here's the deal. I need to find the interval of this derivative function:



f(x)= −5x2+12x−7


So far, I've gotten that the derivative is this:



f-prime(x)= -5x + 12


I tried to make the equation equal to zero. What happened was I got 2.4. Theoretically, anything below 2.4 should be increasing. Everything above 2.4 should be decreasing, I've even tested this.



But the site that I submit it to wants it in interval notation. That's great, but it won't read my interval notation. Am I wrong or is my notation wrong? If I am, what do I do with the derivative?






calculus linear-algebra derivatives factoring







share|cite|improve this question









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    0












    $begingroup$


    Alright, so here's the deal. I need to find the interval of this derivative function:



    f(x)= −5x2+12x−7


    So far, I've gotten that the derivative is this:



    f-prime(x)= -5x + 12


    I tried to make the equation equal to zero. What happened was I got 2.4. Theoretically, anything below 2.4 should be increasing. Everything above 2.4 should be decreasing, I've even tested this.



    But the site that I submit it to wants it in interval notation. That's great, but it won't read my interval notation. Am I wrong or is my notation wrong? If I am, what do I do with the derivative?






    calculus linear-algebra derivatives factoring







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Alright, so here's the deal. I need to find the interval of this derivative function:



      f(x)= −5x2+12x−7


      So far, I've gotten that the derivative is this:



      f-prime(x)= -5x + 12


      I tried to make the equation equal to zero. What happened was I got 2.4. Theoretically, anything below 2.4 should be increasing. Everything above 2.4 should be decreasing, I've even tested this.



      But the site that I submit it to wants it in interval notation. That's great, but it won't read my interval notation. Am I wrong or is my notation wrong? If I am, what do I do with the derivative?






      calculus linear-algebra derivatives factoring







      share|cite|improve this question









      $endgroup$




      Alright, so here's the deal. I need to find the interval of this derivative function:



      f(x)= −5x2+12x−7


      So far, I've gotten that the derivative is this:



      f-prime(x)= -5x + 12


      I tried to make the equation equal to zero. What happened was I got 2.4. Theoretically, anything below 2.4 should be increasing. Everything above 2.4 should be decreasing, I've even tested this.



      But the site that I submit it to wants it in interval notation. That's great, but it won't read my interval notation. Am I wrong or is my notation wrong? If I am, what do I do with the derivative?






      calculus linear-algebra derivatives factoring




      calculus linear-algebra derivatives factoring






      share|cite|improve this question













      share|cite|improve this question











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      asked Oct 22 '14 at 21:05









      user3695903user3695903

      11




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          2 Answers
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          active

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          0












          $begingroup$

          if we have $f(x)=-5x^2+12x-7$ so we get $f'(x)=-10x+12$ and now must be $-10x+12geq 0$ or $-10x+12le 0$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I'm having trouble using either of those equations to find anything other than the same number, which to satisfy both must be equal to x.
            $endgroup$
            – user3695903
            Oct 22 '14 at 21:29










          • $begingroup$
            we have for example $-10x+12geq 0$ or $12geq 10x$ this is equivalent to $frac{12}{10}geq x$ or simplified to $xle frac{6}{5}$
            $endgroup$
            – Dr. Sonnhard Graubner
            Oct 22 '14 at 21:34





















          0












          $begingroup$

          So your first clue should be that it is a function of degree $2$ so it will have only one maximum or minima, in this case, maxima since it is a quadratic opening downward$rightarrow a lt 0$. I am including the reason for this as well. Let us say we have a general quadratic $$ax^2 + bx + c$$ Now taking $x^2$ common we get $$x^2 Bigl(a + frac{b}{x} + frac{c}{x^2}Bigl)$$ Now to determine whether it will open upward or downward we simply see the result of $xrightarrow infty$ and so we get, $$infty^2 Bigl(a + frac{b}{infty} + frac{c}{infty^2}bigl)$$
          So we get, $$infty^2 Bigl(abigl)=0$$
          which obviusly says that if $alt0$ then at $$xrightarrow infty, f(x) rightarrow -infty$$ and for $agt0$ then at $$xrightarrow infty, f(x) rightarrow infty$$ This tells us that this function will be a downward opening parabola as seen in the figure below,
          https://www.desmos.com/calculator/w0krov2wrl

          Now to find the value of the maxima we simply set $$frac{df}{dx} = 0$$
          That is, $$frac{dBigl(-5x^2 + 12x -7bigl)}{dx} = 0$$ Which gives us, $$-10x + 12 = 0$$ Or, $$x = frac{6}{5}$$ Now obviously since this is the maxima it means that after this point the function will be strictly decreasing so the interval for which it is strictly increasing is $$forall ; x lt frac6 5$$






          share|cite|improve this answer











          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

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            active

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            active

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            0












            $begingroup$

            if we have $f(x)=-5x^2+12x-7$ so we get $f'(x)=-10x+12$ and now must be $-10x+12geq 0$ or $-10x+12le 0$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I'm having trouble using either of those equations to find anything other than the same number, which to satisfy both must be equal to x.
              $endgroup$
              – user3695903
              Oct 22 '14 at 21:29










            • $begingroup$
              we have for example $-10x+12geq 0$ or $12geq 10x$ this is equivalent to $frac{12}{10}geq x$ or simplified to $xle frac{6}{5}$
              $endgroup$
              – Dr. Sonnhard Graubner
              Oct 22 '14 at 21:34


















            0












            $begingroup$

            if we have $f(x)=-5x^2+12x-7$ so we get $f'(x)=-10x+12$ and now must be $-10x+12geq 0$ or $-10x+12le 0$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I'm having trouble using either of those equations to find anything other than the same number, which to satisfy both must be equal to x.
              $endgroup$
              – user3695903
              Oct 22 '14 at 21:29










            • $begingroup$
              we have for example $-10x+12geq 0$ or $12geq 10x$ this is equivalent to $frac{12}{10}geq x$ or simplified to $xle frac{6}{5}$
              $endgroup$
              – Dr. Sonnhard Graubner
              Oct 22 '14 at 21:34
















            0












            0








            0





            $begingroup$

            if we have $f(x)=-5x^2+12x-7$ so we get $f'(x)=-10x+12$ and now must be $-10x+12geq 0$ or $-10x+12le 0$






            share|cite|improve this answer









            $endgroup$



            if we have $f(x)=-5x^2+12x-7$ so we get $f'(x)=-10x+12$ and now must be $-10x+12geq 0$ or $-10x+12le 0$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 22 '14 at 21:07









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            78k42866




            78k42866












            • $begingroup$
              I'm having trouble using either of those equations to find anything other than the same number, which to satisfy both must be equal to x.
              $endgroup$
              – user3695903
              Oct 22 '14 at 21:29










            • $begingroup$
              we have for example $-10x+12geq 0$ or $12geq 10x$ this is equivalent to $frac{12}{10}geq x$ or simplified to $xle frac{6}{5}$
              $endgroup$
              – Dr. Sonnhard Graubner
              Oct 22 '14 at 21:34




















            • $begingroup$
              I'm having trouble using either of those equations to find anything other than the same number, which to satisfy both must be equal to x.
              $endgroup$
              – user3695903
              Oct 22 '14 at 21:29










            • $begingroup$
              we have for example $-10x+12geq 0$ or $12geq 10x$ this is equivalent to $frac{12}{10}geq x$ or simplified to $xle frac{6}{5}$
              $endgroup$
              – Dr. Sonnhard Graubner
              Oct 22 '14 at 21:34


















            $begingroup$
            I'm having trouble using either of those equations to find anything other than the same number, which to satisfy both must be equal to x.
            $endgroup$
            – user3695903
            Oct 22 '14 at 21:29




            $begingroup$
            I'm having trouble using either of those equations to find anything other than the same number, which to satisfy both must be equal to x.
            $endgroup$
            – user3695903
            Oct 22 '14 at 21:29












            $begingroup$
            we have for example $-10x+12geq 0$ or $12geq 10x$ this is equivalent to $frac{12}{10}geq x$ or simplified to $xle frac{6}{5}$
            $endgroup$
            – Dr. Sonnhard Graubner
            Oct 22 '14 at 21:34






            $begingroup$
            we have for example $-10x+12geq 0$ or $12geq 10x$ this is equivalent to $frac{12}{10}geq x$ or simplified to $xle frac{6}{5}$
            $endgroup$
            – Dr. Sonnhard Graubner
            Oct 22 '14 at 21:34













            0












            $begingroup$

            So your first clue should be that it is a function of degree $2$ so it will have only one maximum or minima, in this case, maxima since it is a quadratic opening downward$rightarrow a lt 0$. I am including the reason for this as well. Let us say we have a general quadratic $$ax^2 + bx + c$$ Now taking $x^2$ common we get $$x^2 Bigl(a + frac{b}{x} + frac{c}{x^2}Bigl)$$ Now to determine whether it will open upward or downward we simply see the result of $xrightarrow infty$ and so we get, $$infty^2 Bigl(a + frac{b}{infty} + frac{c}{infty^2}bigl)$$
            So we get, $$infty^2 Bigl(abigl)=0$$
            which obviusly says that if $alt0$ then at $$xrightarrow infty, f(x) rightarrow -infty$$ and for $agt0$ then at $$xrightarrow infty, f(x) rightarrow infty$$ This tells us that this function will be a downward opening parabola as seen in the figure below,
            https://www.desmos.com/calculator/w0krov2wrl

            Now to find the value of the maxima we simply set $$frac{df}{dx} = 0$$
            That is, $$frac{dBigl(-5x^2 + 12x -7bigl)}{dx} = 0$$ Which gives us, $$-10x + 12 = 0$$ Or, $$x = frac{6}{5}$$ Now obviously since this is the maxima it means that after this point the function will be strictly decreasing so the interval for which it is strictly increasing is $$forall ; x lt frac6 5$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              So your first clue should be that it is a function of degree $2$ so it will have only one maximum or minima, in this case, maxima since it is a quadratic opening downward$rightarrow a lt 0$. I am including the reason for this as well. Let us say we have a general quadratic $$ax^2 + bx + c$$ Now taking $x^2$ common we get $$x^2 Bigl(a + frac{b}{x} + frac{c}{x^2}Bigl)$$ Now to determine whether it will open upward or downward we simply see the result of $xrightarrow infty$ and so we get, $$infty^2 Bigl(a + frac{b}{infty} + frac{c}{infty^2}bigl)$$
              So we get, $$infty^2 Bigl(abigl)=0$$
              which obviusly says that if $alt0$ then at $$xrightarrow infty, f(x) rightarrow -infty$$ and for $agt0$ then at $$xrightarrow infty, f(x) rightarrow infty$$ This tells us that this function will be a downward opening parabola as seen in the figure below,
              https://www.desmos.com/calculator/w0krov2wrl

              Now to find the value of the maxima we simply set $$frac{df}{dx} = 0$$
              That is, $$frac{dBigl(-5x^2 + 12x -7bigl)}{dx} = 0$$ Which gives us, $$-10x + 12 = 0$$ Or, $$x = frac{6}{5}$$ Now obviously since this is the maxima it means that after this point the function will be strictly decreasing so the interval for which it is strictly increasing is $$forall ; x lt frac6 5$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                So your first clue should be that it is a function of degree $2$ so it will have only one maximum or minima, in this case, maxima since it is a quadratic opening downward$rightarrow a lt 0$. I am including the reason for this as well. Let us say we have a general quadratic $$ax^2 + bx + c$$ Now taking $x^2$ common we get $$x^2 Bigl(a + frac{b}{x} + frac{c}{x^2}Bigl)$$ Now to determine whether it will open upward or downward we simply see the result of $xrightarrow infty$ and so we get, $$infty^2 Bigl(a + frac{b}{infty} + frac{c}{infty^2}bigl)$$
                So we get, $$infty^2 Bigl(abigl)=0$$
                which obviusly says that if $alt0$ then at $$xrightarrow infty, f(x) rightarrow -infty$$ and for $agt0$ then at $$xrightarrow infty, f(x) rightarrow infty$$ This tells us that this function will be a downward opening parabola as seen in the figure below,
                https://www.desmos.com/calculator/w0krov2wrl

                Now to find the value of the maxima we simply set $$frac{df}{dx} = 0$$
                That is, $$frac{dBigl(-5x^2 + 12x -7bigl)}{dx} = 0$$ Which gives us, $$-10x + 12 = 0$$ Or, $$x = frac{6}{5}$$ Now obviously since this is the maxima it means that after this point the function will be strictly decreasing so the interval for which it is strictly increasing is $$forall ; x lt frac6 5$$






                share|cite|improve this answer











                $endgroup$



                So your first clue should be that it is a function of degree $2$ so it will have only one maximum or minima, in this case, maxima since it is a quadratic opening downward$rightarrow a lt 0$. I am including the reason for this as well. Let us say we have a general quadratic $$ax^2 + bx + c$$ Now taking $x^2$ common we get $$x^2 Bigl(a + frac{b}{x} + frac{c}{x^2}Bigl)$$ Now to determine whether it will open upward or downward we simply see the result of $xrightarrow infty$ and so we get, $$infty^2 Bigl(a + frac{b}{infty} + frac{c}{infty^2}bigl)$$
                So we get, $$infty^2 Bigl(abigl)=0$$
                which obviusly says that if $alt0$ then at $$xrightarrow infty, f(x) rightarrow -infty$$ and for $agt0$ then at $$xrightarrow infty, f(x) rightarrow infty$$ This tells us that this function will be a downward opening parabola as seen in the figure below,
                https://www.desmos.com/calculator/w0krov2wrl

                Now to find the value of the maxima we simply set $$frac{df}{dx} = 0$$
                That is, $$frac{dBigl(-5x^2 + 12x -7bigl)}{dx} = 0$$ Which gives us, $$-10x + 12 = 0$$ Or, $$x = frac{6}{5}$$ Now obviously since this is the maxima it means that after this point the function will be strictly decreasing so the interval for which it is strictly increasing is $$forall ; x lt frac6 5$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 17 '18 at 7:26

























                answered Nov 17 '18 at 7:14









                Prakhar NagpalPrakhar Nagpal

                752318




                752318






























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