Finding absolute and relative extrema of a function











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I don't know if what I'm doing is correct.



Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$.
Choose the correct answer:





  1. $f$ has an absolute maximum at $x=0$.


  2. $f$ has a relative maximum at $x=0$ which isn't absolute.


  3. $f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)


  4. $f$ doesn't have relative maximum.


What I've been doing:



Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.



I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).



When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $lim_{x to +infty} f(x) = +infty$ (is finding the limit as $x to +infty$ correct when trying to find the absolute extrema?).



So, how do I prove that option 2 is correct without looking at the graphic?






calculus real-analysis







share|cite|improve this question






















  • I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
    – José Carlos Santos
    Nov 22 at 16:49












  • Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
    – parishilton
    Nov 22 at 16:52

















up vote
0
down vote

favorite












I don't know if what I'm doing is correct.



Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$.
Choose the correct answer:





  1. $f$ has an absolute maximum at $x=0$.


  2. $f$ has a relative maximum at $x=0$ which isn't absolute.


  3. $f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)


  4. $f$ doesn't have relative maximum.


What I've been doing:



Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.



I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).



When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $lim_{x to +infty} f(x) = +infty$ (is finding the limit as $x to +infty$ correct when trying to find the absolute extrema?).



So, how do I prove that option 2 is correct without looking at the graphic?






calculus real-analysis







share|cite|improve this question






















  • I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
    – José Carlos Santos
    Nov 22 at 16:49












  • Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
    – parishilton
    Nov 22 at 16:52















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I don't know if what I'm doing is correct.



Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$.
Choose the correct answer:





  1. $f$ has an absolute maximum at $x=0$.


  2. $f$ has a relative maximum at $x=0$ which isn't absolute.


  3. $f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)


  4. $f$ doesn't have relative maximum.


What I've been doing:



Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.



I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).



When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $lim_{x to +infty} f(x) = +infty$ (is finding the limit as $x to +infty$ correct when trying to find the absolute extrema?).



So, how do I prove that option 2 is correct without looking at the graphic?






calculus real-analysis







share|cite|improve this question













I don't know if what I'm doing is correct.



Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$.
Choose the correct answer:





  1. $f$ has an absolute maximum at $x=0$.


  2. $f$ has a relative maximum at $x=0$ which isn't absolute.


  3. $f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)


  4. $f$ doesn't have relative maximum.


What I've been doing:



Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.



I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).



When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $lim_{x to +infty} f(x) = +infty$ (is finding the limit as $x to +infty$ correct when trying to find the absolute extrema?).



So, how do I prove that option 2 is correct without looking at the graphic?






calculus real-analysis




calculus real-analysis






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asked Nov 22 at 16:46









parishilton

15910




15910












  • I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
    – José Carlos Santos
    Nov 22 at 16:49












  • Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
    – parishilton
    Nov 22 at 16:52




















  • I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
    – José Carlos Santos
    Nov 22 at 16:49












  • Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
    – parishilton
    Nov 22 at 16:52


















I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 at 16:49






I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 at 16:49














Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 at 16:52






Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 at 16:52












1 Answer
1






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1
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HINT



We have that



$$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$



then consider the sign of $f'(x)$.






share|cite|improve this answer





















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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    HINT



    We have that



    $$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$



    then consider the sign of $f'(x)$.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      HINT



      We have that



      $$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$



      then consider the sign of $f'(x)$.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        HINT



        We have that



        $$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$



        then consider the sign of $f'(x)$.






        share|cite|improve this answer












        HINT



        We have that



        $$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$



        then consider the sign of $f'(x)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 16:49









        gimusi

        92.7k94495




        92.7k94495






























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