Unique Line Tangent to 2 points on the curve y=x^4+x^3

jonbenedick shared this question 7 years ago
Answered

How do you draw a unique line tangent to the two points on the curve y=x^4+x^3

Comments (7)

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Do you mean like this, or something else?

http://tube.geogebra.org/ma...

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hello

i got with Maxima that you must use the point A=((sqrt(3)-1)/4,0) and f(x)=x^4+x^3 then Tangent[A, f]

i can not get it with CAS because i do not know how to say to CAS a!=c for solutions of equations


saludos86efb4b78d37a9fb9c24fb52ac45e8fd

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hello

it is possible with CAS substituting by hand for avoiding 0/0


saludos4e093e5638ec9f6d14ae1c44411dc28c

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The common tangent slope is the same of the tangent at point A(-1/4,f(-1/4)), being -1/4 the solution of equation: third derivative of f(x)=0. See att'd figure.8775c4bf1accbf2089d940c59ca4a471

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Finalized.e727f4dc41d3f2f1d0eab7f622919732

Files: ct.jpg
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More details.7956a076b761e934a6d83a9a8d73f6c2

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MatLab solution, according to mathmagic's scheme.

    % ------------- twovars2.m ------------- (May.20,2015) ------

    % ----- Successive approx. 2x2 system of NON LINEAR equations ------

    % --------------------------------------------------------------------------

    clc, clear, close all % clear screen

    format long

    % -------------------------------------------------

    % 2 starting values of iteration

    a=0.8; c=-1.2; % a= right-hand abscissa; c= left-hand abscissa

    %

    % System to solve: F1(a,c)=0; F2(a,c)=0

    while 1 % Start iteration

    F(1)= (4*a^3 + 3*a^2)*(a-c) - (a^4+a^3-c^4-c^3); % F1

    F(2)= (4*c^3 + 3*c^2)*(a-c) - (a^4+a^3-c^4-c^3); % F2


    % ------- W = jacobian matrix ----------

    W(1,1)= (12*a^2 + 6*a)*(a-c)+ (4*a^3 + 3*a^2) -4*a^3-3*a^2; % F11= dF1/da |F12 F12|

    W(1,2)= -(4*a^3 + 3*a^2)+4*c^3+3*c^2; % F12= dF1/dc W= | |

    % |F21 F22|

    W(2,1)= 4*c^3 + 3*c^2-4*a^3-3*a^2; % F21= dF2/da

    W(2,2)= (12*c^2+6*c)*(a-c)-(4*c^3 + 3*c^2) +4*c^3+3*c^2; % F22= dF2/dc

    %

    P=inv(W)*F'; % Xi+1= Xi - inv(W)*F(Xi); iteration vars Xi=(ai,ci)

    %

    a1 = a - P(1); c1 = c - P(2);

    fprintf ('\n ai = %9.6f ci = %9.6f', a1,c1);


    if abs(P(1)) < 1e-8 & abs(P(2)) < 1e-8

    break;

    end

    a = a1; c = c1;


    end; % end while 1 (iteration)

    %

    fprintf ('\n\n RESULTS \n a = %9.6f c = %9.6f \n', a1,c1);

    %

    % ------- EOF: twovars2.m ---------

    % Output, convergence per successive approximations

    % ai = 0.559666 ci = -0.992490

    % ai = 0.388581 ci = -0.847549

    % ai = 0.274385 ci = -0.753663

    % ai = 0.209812 ci = -0.702834

    % ai = 0.186168 ci = -0.685221

    % ai = 0.183063 ci = -0.683046

    % ai = 0.183013 ci = -0.683013

    % ai = 0.183013 ci = -0.683013

    % ai = 0.183013 ci = -0.683013

    %

    % RESULTS

    % a = 0.183013 c = -0.683013

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