How to do this Linear Approximation?












0












$begingroup$


this question has been giving me a little trouble:



Use a linear approximation to estimate the number $8.07^{2/3}$



I tried using $f(a)+f'(a)(x-a)$ but the answer I get ($4.02$) is apparently wrong. Any help would be appreciated!






calculus linear-approximation







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$endgroup$












  • $begingroup$
    What did you use for $f,x,$ and $a$?
    $endgroup$
    – pwerth
    Dec 23 '18 at 3:21










  • $begingroup$
    Maybe you need one more digit after the decimal point
    $endgroup$
    – Andrei
    Dec 23 '18 at 3:21










  • $begingroup$
    @pwerth I used f: x^(2/3), a:8.07
    $endgroup$
    – vmahajan17
    Dec 23 '18 at 3:27










  • $begingroup$
    @Andrei that didn't work either
    $endgroup$
    – vmahajan17
    Dec 23 '18 at 3:27






  • 2




    $begingroup$
    $a=8$, then $8^{2/3}=4$
    $endgroup$
    – Andrei
    Dec 23 '18 at 3:38
















0












$begingroup$


this question has been giving me a little trouble:



Use a linear approximation to estimate the number $8.07^{2/3}$



I tried using $f(a)+f'(a)(x-a)$ but the answer I get ($4.02$) is apparently wrong. Any help would be appreciated!






calculus linear-approximation







share|cite|improve this question











$endgroup$












  • $begingroup$
    What did you use for $f,x,$ and $a$?
    $endgroup$
    – pwerth
    Dec 23 '18 at 3:21










  • $begingroup$
    Maybe you need one more digit after the decimal point
    $endgroup$
    – Andrei
    Dec 23 '18 at 3:21










  • $begingroup$
    @pwerth I used f: x^(2/3), a:8.07
    $endgroup$
    – vmahajan17
    Dec 23 '18 at 3:27










  • $begingroup$
    @Andrei that didn't work either
    $endgroup$
    – vmahajan17
    Dec 23 '18 at 3:27






  • 2




    $begingroup$
    $a=8$, then $8^{2/3}=4$
    $endgroup$
    – Andrei
    Dec 23 '18 at 3:38














0












0








0





$begingroup$


this question has been giving me a little trouble:



Use a linear approximation to estimate the number $8.07^{2/3}$



I tried using $f(a)+f'(a)(x-a)$ but the answer I get ($4.02$) is apparently wrong. Any help would be appreciated!






calculus linear-approximation







share|cite|improve this question











$endgroup$




this question has been giving me a little trouble:



Use a linear approximation to estimate the number $8.07^{2/3}$



I tried using $f(a)+f'(a)(x-a)$ but the answer I get ($4.02$) is apparently wrong. Any help would be appreciated!






calculus linear-approximation




calculus linear-approximation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 3:41









pwerth

3,113417




3,113417










asked Dec 23 '18 at 3:18









vmahajan17vmahajan17

65




65












  • $begingroup$
    What did you use for $f,x,$ and $a$?
    $endgroup$
    – pwerth
    Dec 23 '18 at 3:21










  • $begingroup$
    Maybe you need one more digit after the decimal point
    $endgroup$
    – Andrei
    Dec 23 '18 at 3:21










  • $begingroup$
    @pwerth I used f: x^(2/3), a:8.07
    $endgroup$
    – vmahajan17
    Dec 23 '18 at 3:27










  • $begingroup$
    @Andrei that didn't work either
    $endgroup$
    – vmahajan17
    Dec 23 '18 at 3:27






  • 2




    $begingroup$
    $a=8$, then $8^{2/3}=4$
    $endgroup$
    – Andrei
    Dec 23 '18 at 3:38


















  • $begingroup$
    What did you use for $f,x,$ and $a$?
    $endgroup$
    – pwerth
    Dec 23 '18 at 3:21










  • $begingroup$
    Maybe you need one more digit after the decimal point
    $endgroup$
    – Andrei
    Dec 23 '18 at 3:21










  • $begingroup$
    @pwerth I used f: x^(2/3), a:8.07
    $endgroup$
    – vmahajan17
    Dec 23 '18 at 3:27










  • $begingroup$
    @Andrei that didn't work either
    $endgroup$
    – vmahajan17
    Dec 23 '18 at 3:27






  • 2




    $begingroup$
    $a=8$, then $8^{2/3}=4$
    $endgroup$
    – Andrei
    Dec 23 '18 at 3:38
















$begingroup$
What did you use for $f,x,$ and $a$?
$endgroup$
– pwerth
Dec 23 '18 at 3:21




$begingroup$
What did you use for $f,x,$ and $a$?
$endgroup$
– pwerth
Dec 23 '18 at 3:21












$begingroup$
Maybe you need one more digit after the decimal point
$endgroup$
– Andrei
Dec 23 '18 at 3:21




$begingroup$
Maybe you need one more digit after the decimal point
$endgroup$
– Andrei
Dec 23 '18 at 3:21












$begingroup$
@pwerth I used f: x^(2/3), a:8.07
$endgroup$
– vmahajan17
Dec 23 '18 at 3:27




$begingroup$
@pwerth I used f: x^(2/3), a:8.07
$endgroup$
– vmahajan17
Dec 23 '18 at 3:27












$begingroup$
@Andrei that didn't work either
$endgroup$
– vmahajan17
Dec 23 '18 at 3:27




$begingroup$
@Andrei that didn't work either
$endgroup$
– vmahajan17
Dec 23 '18 at 3:27




2




2




$begingroup$
$a=8$, then $8^{2/3}=4$
$endgroup$
– Andrei
Dec 23 '18 at 3:38




$begingroup$
$a=8$, then $8^{2/3}=4$
$endgroup$
– Andrei
Dec 23 '18 at 3:38










2 Answers
2






active

oldest

votes


















1












$begingroup$

If you use $f(x)=x^{2/3}$, you have $f(x)approx f(a)+f'(a)(x-a)$. $f'(x)=frac 23 x^{-1/3}$. If you plug in $a=8$, $f'(8)=frac 13$, so $f(8.07)=4+0.07/3=4.02333$. The real answer is $4.023299$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Take $f(x)=x^{2/3}$ and $a=8$. Then
    $$f(x)=x^{2/3} Rightarrow f'(x)=frac{2}{3}x^{-1/3} Rightarrow f'(a)=frac{2}{3}cdot 8^{-1/3}=frac{1}{3}$$



    So $f(a)+f'(a)(x-a)=f(8)+frac{1}{3}(.07)=4+frac{1}{3}(.07)approx4.023$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      should be $x^{-1/3}$
      $endgroup$
      – Andrei
      Dec 23 '18 at 3:44










    • $begingroup$
      @Andrei Yep, thanks
      $endgroup$
      – pwerth
      Dec 23 '18 at 3:45











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    2 Answers
    2






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    2 Answers
    2






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    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    If you use $f(x)=x^{2/3}$, you have $f(x)approx f(a)+f'(a)(x-a)$. $f'(x)=frac 23 x^{-1/3}$. If you plug in $a=8$, $f'(8)=frac 13$, so $f(8.07)=4+0.07/3=4.02333$. The real answer is $4.023299$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If you use $f(x)=x^{2/3}$, you have $f(x)approx f(a)+f'(a)(x-a)$. $f'(x)=frac 23 x^{-1/3}$. If you plug in $a=8$, $f'(8)=frac 13$, so $f(8.07)=4+0.07/3=4.02333$. The real answer is $4.023299$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If you use $f(x)=x^{2/3}$, you have $f(x)approx f(a)+f'(a)(x-a)$. $f'(x)=frac 23 x^{-1/3}$. If you plug in $a=8$, $f'(8)=frac 13$, so $f(8.07)=4+0.07/3=4.02333$. The real answer is $4.023299$






        share|cite|improve this answer









        $endgroup$



        If you use $f(x)=x^{2/3}$, you have $f(x)approx f(a)+f'(a)(x-a)$. $f'(x)=frac 23 x^{-1/3}$. If you plug in $a=8$, $f'(8)=frac 13$, so $f(8.07)=4+0.07/3=4.02333$. The real answer is $4.023299$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 23 '18 at 3:43









        AndreiAndrei

        11.9k21126




        11.9k21126























            1












            $begingroup$

            Take $f(x)=x^{2/3}$ and $a=8$. Then
            $$f(x)=x^{2/3} Rightarrow f'(x)=frac{2}{3}x^{-1/3} Rightarrow f'(a)=frac{2}{3}cdot 8^{-1/3}=frac{1}{3}$$



            So $f(a)+f'(a)(x-a)=f(8)+frac{1}{3}(.07)=4+frac{1}{3}(.07)approx4.023$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              should be $x^{-1/3}$
              $endgroup$
              – Andrei
              Dec 23 '18 at 3:44










            • $begingroup$
              @Andrei Yep, thanks
              $endgroup$
              – pwerth
              Dec 23 '18 at 3:45
















            1












            $begingroup$

            Take $f(x)=x^{2/3}$ and $a=8$. Then
            $$f(x)=x^{2/3} Rightarrow f'(x)=frac{2}{3}x^{-1/3} Rightarrow f'(a)=frac{2}{3}cdot 8^{-1/3}=frac{1}{3}$$



            So $f(a)+f'(a)(x-a)=f(8)+frac{1}{3}(.07)=4+frac{1}{3}(.07)approx4.023$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              should be $x^{-1/3}$
              $endgroup$
              – Andrei
              Dec 23 '18 at 3:44










            • $begingroup$
              @Andrei Yep, thanks
              $endgroup$
              – pwerth
              Dec 23 '18 at 3:45














            1












            1








            1





            $begingroup$

            Take $f(x)=x^{2/3}$ and $a=8$. Then
            $$f(x)=x^{2/3} Rightarrow f'(x)=frac{2}{3}x^{-1/3} Rightarrow f'(a)=frac{2}{3}cdot 8^{-1/3}=frac{1}{3}$$



            So $f(a)+f'(a)(x-a)=f(8)+frac{1}{3}(.07)=4+frac{1}{3}(.07)approx4.023$






            share|cite|improve this answer











            $endgroup$



            Take $f(x)=x^{2/3}$ and $a=8$. Then
            $$f(x)=x^{2/3} Rightarrow f'(x)=frac{2}{3}x^{-1/3} Rightarrow f'(a)=frac{2}{3}cdot 8^{-1/3}=frac{1}{3}$$



            So $f(a)+f'(a)(x-a)=f(8)+frac{1}{3}(.07)=4+frac{1}{3}(.07)approx4.023$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 23 '18 at 3:45

























            answered Dec 23 '18 at 3:42









            pwerthpwerth

            3,113417




            3,113417












            • $begingroup$
              should be $x^{-1/3}$
              $endgroup$
              – Andrei
              Dec 23 '18 at 3:44










            • $begingroup$
              @Andrei Yep, thanks
              $endgroup$
              – pwerth
              Dec 23 '18 at 3:45


















            • $begingroup$
              should be $x^{-1/3}$
              $endgroup$
              – Andrei
              Dec 23 '18 at 3:44










            • $begingroup$
              @Andrei Yep, thanks
              $endgroup$
              – pwerth
              Dec 23 '18 at 3:45
















            $begingroup$
            should be $x^{-1/3}$
            $endgroup$
            – Andrei
            Dec 23 '18 at 3:44




            $begingroup$
            should be $x^{-1/3}$
            $endgroup$
            – Andrei
            Dec 23 '18 at 3:44












            $begingroup$
            @Andrei Yep, thanks
            $endgroup$
            – pwerth
            Dec 23 '18 at 3:45




            $begingroup$
            @Andrei Yep, thanks
            $endgroup$
            – pwerth
            Dec 23 '18 at 3:45


















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