Find the tangent line of $y=x-e^{-x}$ parallel to $6x-2y=7$












-1












$begingroup$


Find the tangent line of $y=x-e^{-x}$ parallel to $6x-2y=7$



I know I have to take the derivative of the first equation, but I don’t know what to do after.






calculus







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  • $begingroup$
    You want to know where the derivative of both equations match.
    $endgroup$
    – Kaynex
    Jan 14 at 4:36










  • $begingroup$
    What does it mean mathematically for a line to be 'parallel' to a different line?
    $endgroup$
    – coreyman317
    Jan 14 at 4:37
















-1












$begingroup$


Find the tangent line of $y=x-e^{-x}$ parallel to $6x-2y=7$



I know I have to take the derivative of the first equation, but I don’t know what to do after.






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You want to know where the derivative of both equations match.
    $endgroup$
    – Kaynex
    Jan 14 at 4:36










  • $begingroup$
    What does it mean mathematically for a line to be 'parallel' to a different line?
    $endgroup$
    – coreyman317
    Jan 14 at 4:37














-1












-1








-1





$begingroup$


Find the tangent line of $y=x-e^{-x}$ parallel to $6x-2y=7$



I know I have to take the derivative of the first equation, but I don’t know what to do after.






calculus







share|cite|improve this question











$endgroup$




Find the tangent line of $y=x-e^{-x}$ parallel to $6x-2y=7$



I know I have to take the derivative of the first equation, but I don’t know what to do after.






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 4:47









coreyman317

732420




732420










asked Jan 14 at 4:34









Tara AllahdadiTara Allahdadi

1




1












  • $begingroup$
    You want to know where the derivative of both equations match.
    $endgroup$
    – Kaynex
    Jan 14 at 4:36










  • $begingroup$
    What does it mean mathematically for a line to be 'parallel' to a different line?
    $endgroup$
    – coreyman317
    Jan 14 at 4:37


















  • $begingroup$
    You want to know where the derivative of both equations match.
    $endgroup$
    – Kaynex
    Jan 14 at 4:36










  • $begingroup$
    What does it mean mathematically for a line to be 'parallel' to a different line?
    $endgroup$
    – coreyman317
    Jan 14 at 4:37
















$begingroup$
You want to know where the derivative of both equations match.
$endgroup$
– Kaynex
Jan 14 at 4:36




$begingroup$
You want to know where the derivative of both equations match.
$endgroup$
– Kaynex
Jan 14 at 4:36












$begingroup$
What does it mean mathematically for a line to be 'parallel' to a different line?
$endgroup$
– coreyman317
Jan 14 at 4:37




$begingroup$
What does it mean mathematically for a line to be 'parallel' to a different line?
$endgroup$
– coreyman317
Jan 14 at 4:37










3 Answers
3






active

oldest

votes


















1












$begingroup$

For two lines to be parallel, their slopes must equal. You can compute the slope of the tangent line of $x-e^x$ by taking the derivative. For the line L, put it in slope-intercept form. Call that slope m. Then you want to solve $f^prime$(x) = m. Those are the x-values of the points on the graph of f(x) parallel to L.
Then you can use the point-slope formula to find the equation of the tangent lines.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The line $6x-2y = 7$ has slope $3$ thus, the tangent line to $y = x - e^{-x}$ that is parallel occurs when $1 + e^{-x} = 3$ or when $e^{-x} = 2 $ which means that $x = - ln 2 $.



    So you know you need to find the equation of the tangent line at the point $(- ln 2, -2 - ln 2 )$. Can you finish it?






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      The tangent line of a function $f(x)$ at $(x_0,y_0)$ has the following form: $$y-y_0=f'(x_0)(x-x_0)$$



      We want to find a tangent line of $f(x)=x-e^{-x}$ that is parallel to $6x-2y=7$ or $y=3x-frac{7}{2}$. All that is required for a line to be parallel to another line is their two slopes be equal. Hence we need $$f'(x_0)=3implies(x-e^{-x})^{'}=3implies1+e^{-x}=3implies e^{-x}=2 $$ $$impliesln(e^{-x})=ln(2)implies-x=ln(2)spacetext{or}space x=-ln(2)$$ We want the tangent line of $f(x)=x-e^{-x}$ at $x=-ln(2)$.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        For two lines to be parallel, their slopes must equal. You can compute the slope of the tangent line of $x-e^x$ by taking the derivative. For the line L, put it in slope-intercept form. Call that slope m. Then you want to solve $f^prime$(x) = m. Those are the x-values of the points on the graph of f(x) parallel to L.
        Then you can use the point-slope formula to find the equation of the tangent lines.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          For two lines to be parallel, their slopes must equal. You can compute the slope of the tangent line of $x-e^x$ by taking the derivative. For the line L, put it in slope-intercept form. Call that slope m. Then you want to solve $f^prime$(x) = m. Those are the x-values of the points on the graph of f(x) parallel to L.
          Then you can use the point-slope formula to find the equation of the tangent lines.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            For two lines to be parallel, their slopes must equal. You can compute the slope of the tangent line of $x-e^x$ by taking the derivative. For the line L, put it in slope-intercept form. Call that slope m. Then you want to solve $f^prime$(x) = m. Those are the x-values of the points on the graph of f(x) parallel to L.
            Then you can use the point-slope formula to find the equation of the tangent lines.






            share|cite|improve this answer









            $endgroup$



            For two lines to be parallel, their slopes must equal. You can compute the slope of the tangent line of $x-e^x$ by taking the derivative. For the line L, put it in slope-intercept form. Call that slope m. Then you want to solve $f^prime$(x) = m. Those are the x-values of the points on the graph of f(x) parallel to L.
            Then you can use the point-slope formula to find the equation of the tangent lines.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 14 at 4:40









            Joel PereiraJoel Pereira

            74519




            74519























                1












                $begingroup$

                The line $6x-2y = 7$ has slope $3$ thus, the tangent line to $y = x - e^{-x}$ that is parallel occurs when $1 + e^{-x} = 3$ or when $e^{-x} = 2 $ which means that $x = - ln 2 $.



                So you know you need to find the equation of the tangent line at the point $(- ln 2, -2 - ln 2 )$. Can you finish it?






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  The line $6x-2y = 7$ has slope $3$ thus, the tangent line to $y = x - e^{-x}$ that is parallel occurs when $1 + e^{-x} = 3$ or when $e^{-x} = 2 $ which means that $x = - ln 2 $.



                  So you know you need to find the equation of the tangent line at the point $(- ln 2, -2 - ln 2 )$. Can you finish it?






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    The line $6x-2y = 7$ has slope $3$ thus, the tangent line to $y = x - e^{-x}$ that is parallel occurs when $1 + e^{-x} = 3$ or when $e^{-x} = 2 $ which means that $x = - ln 2 $.



                    So you know you need to find the equation of the tangent line at the point $(- ln 2, -2 - ln 2 )$. Can you finish it?






                    share|cite|improve this answer











                    $endgroup$



                    The line $6x-2y = 7$ has slope $3$ thus, the tangent line to $y = x - e^{-x}$ that is parallel occurs when $1 + e^{-x} = 3$ or when $e^{-x} = 2 $ which means that $x = - ln 2 $.



                    So you know you need to find the equation of the tangent line at the point $(- ln 2, -2 - ln 2 )$. Can you finish it?







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 14 at 4:42

























                    answered Jan 14 at 4:40









                    Jimmy SabaterJimmy Sabater

                    2,640321




                    2,640321























                        0












                        $begingroup$

                        The tangent line of a function $f(x)$ at $(x_0,y_0)$ has the following form: $$y-y_0=f'(x_0)(x-x_0)$$



                        We want to find a tangent line of $f(x)=x-e^{-x}$ that is parallel to $6x-2y=7$ or $y=3x-frac{7}{2}$. All that is required for a line to be parallel to another line is their two slopes be equal. Hence we need $$f'(x_0)=3implies(x-e^{-x})^{'}=3implies1+e^{-x}=3implies e^{-x}=2 $$ $$impliesln(e^{-x})=ln(2)implies-x=ln(2)spacetext{or}space x=-ln(2)$$ We want the tangent line of $f(x)=x-e^{-x}$ at $x=-ln(2)$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The tangent line of a function $f(x)$ at $(x_0,y_0)$ has the following form: $$y-y_0=f'(x_0)(x-x_0)$$



                          We want to find a tangent line of $f(x)=x-e^{-x}$ that is parallel to $6x-2y=7$ or $y=3x-frac{7}{2}$. All that is required for a line to be parallel to another line is their two slopes be equal. Hence we need $$f'(x_0)=3implies(x-e^{-x})^{'}=3implies1+e^{-x}=3implies e^{-x}=2 $$ $$impliesln(e^{-x})=ln(2)implies-x=ln(2)spacetext{or}space x=-ln(2)$$ We want the tangent line of $f(x)=x-e^{-x}$ at $x=-ln(2)$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The tangent line of a function $f(x)$ at $(x_0,y_0)$ has the following form: $$y-y_0=f'(x_0)(x-x_0)$$



                            We want to find a tangent line of $f(x)=x-e^{-x}$ that is parallel to $6x-2y=7$ or $y=3x-frac{7}{2}$. All that is required for a line to be parallel to another line is their two slopes be equal. Hence we need $$f'(x_0)=3implies(x-e^{-x})^{'}=3implies1+e^{-x}=3implies e^{-x}=2 $$ $$impliesln(e^{-x})=ln(2)implies-x=ln(2)spacetext{or}space x=-ln(2)$$ We want the tangent line of $f(x)=x-e^{-x}$ at $x=-ln(2)$.






                            share|cite|improve this answer









                            $endgroup$



                            The tangent line of a function $f(x)$ at $(x_0,y_0)$ has the following form: $$y-y_0=f'(x_0)(x-x_0)$$



                            We want to find a tangent line of $f(x)=x-e^{-x}$ that is parallel to $6x-2y=7$ or $y=3x-frac{7}{2}$. All that is required for a line to be parallel to another line is their two slopes be equal. Hence we need $$f'(x_0)=3implies(x-e^{-x})^{'}=3implies1+e^{-x}=3implies e^{-x}=2 $$ $$impliesln(e^{-x})=ln(2)implies-x=ln(2)spacetext{or}space x=-ln(2)$$ We want the tangent line of $f(x)=x-e^{-x}$ at $x=-ln(2)$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 14 at 4:52









                            coreyman317coreyman317

                            732420




                            732420






























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