Reversing sign of third derivative

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I don't understand some part of the solution given to this question:



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I can understand how the sign of $f''(x)$ can be reversed (i.e. flipping over the x-axis) but I don't get why changing $x$ to $-x$ would change the sign of $f'''(x)$?




calculus derivatives




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    up vote
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    I don't understand some part of the solution given to this question:



    enter image description here



    I can understand how the sign of $f''(x)$ can be reversed (i.e. flipping over the x-axis) but I don't get why changing $x$ to $-x$ would change the sign of $f'''(x)$?




    calculus derivatives




    share|cite|improve this question





















      up vote
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      favorite
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      up vote
      2
      down vote

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      1





      I don't understand some part of the solution given to this question:



      enter image description here



      I can understand how the sign of $f''(x)$ can be reversed (i.e. flipping over the x-axis) but I don't get why changing $x$ to $-x$ would change the sign of $f'''(x)$?




      calculus derivatives




      share|cite|improve this question











      I don't understand some part of the solution given to this question:



      enter image description here



      I can understand how the sign of $f''(x)$ can be reversed (i.e. flipping over the x-axis) but I don't get why changing $x$ to $-x$ would change the sign of $f'''(x)$?





      calculus derivatives





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          Let $g(x)=f(-x)$. Then using the chain rule in each step, we get $$g'(x)=-f'(-x),$$ $$g''(x)=-(-f''(-x))=f''(-x),$$ and finally $$g'''(x)=-f'''(-x).$$ So if $f'''$ is always positive then $g'''$ is always negative, and vice versa.






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            Let $g(x)=f(-x)$. Then using the chain rule in each step, we get $$g'(x)=-f'(-x),$$ $$g''(x)=-(-f''(-x))=f''(-x),$$ and finally $$g'''(x)=-f'''(-x).$$ So if $f'''$ is always positive then $g'''$ is always negative, and vice versa.






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              up vote
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              Let $g(x)=f(-x)$. Then using the chain rule in each step, we get $$g'(x)=-f'(-x),$$ $$g''(x)=-(-f''(-x))=f''(-x),$$ and finally $$g'''(x)=-f'''(-x).$$ So if $f'''$ is always positive then $g'''$ is always negative, and vice versa.






              share|cite|improve this answer























                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                Let $g(x)=f(-x)$. Then using the chain rule in each step, we get $$g'(x)=-f'(-x),$$ $$g''(x)=-(-f''(-x))=f''(-x),$$ and finally $$g'''(x)=-f'''(-x).$$ So if $f'''$ is always positive then $g'''$ is always negative, and vice versa.






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                Let $g(x)=f(-x)$. Then using the chain rule in each step, we get $$g'(x)=-f'(-x),$$ $$g''(x)=-(-f''(-x))=f''(-x),$$ and finally $$g'''(x)=-f'''(-x).$$ So if $f'''$ is always positive then $g'''$ is always negative, and vice versa.







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                answered Aug 4 at 3:26









                Eric Wofsey

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