Calculus I

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  • #13253

    Ezra Halleck
    Participant

    6. 3.8: Calculate b and where is inverse of . Find the tangent lines to f at x=1 and to g at x=b.
    a.
    b.
    (Use the inverse function theorem: If , then )

    #16857

    Ezra Halleck
    Participant

    a. We find f ‘ (x)= 6x^2-6x+6 and evaluate at g(b)=1 to get 6-6+6=6
    so g'(b)=1/6
    b=f(1)=2-3+6+1=6
    Tangent line to f is y-6=6(x-1)
    Tangent line to g is y-1=1/6(x-6)
    b. Rewrite f(x)= (x^3+1)* (x^2+1)^(-1/2). We find f ‘ (x) using the product rule:
    (3x^2)(x^2+1)^(-1/2)+(x^3+1)(-1/2)(x^2+1)^(-3/2)(2x) and evaluate at 1:
    f ‘ (1)=3/sqrt(2)-1/sqrt(2)=2/sqrt(2))=sqrt(2)
    so g ‘ (b)= 1/sqrt(2)
    b=f(1)=2/sqrt(2)=sqrt(2)
    Tangent line to f is y-sqrt(2)=sqrt(2)(x-1)
    Tangent line to g is y-1=1/sqrt(2)(x-sqrt(2))

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