14a) find the slope of the tangent to the curve a...

Answers ( 3 )

  1. julia williams
    julia williams
    so the slope at the point where x = a is
    solpe at the point (1,1) is = -1/2
    so the equation of the tangent line is (y-1) =(-1/2)(x-1)
    y = -x/2 +3/2
    slope at the point (4,1/2) is = -1/16
    then the equation of the tangent line at ( 4,1/2) is


    y = -x/16+3/4
  2. camille ennis
    camille ennis
    Quotient rule: (f/g)'=(gf'-fg')/g^2
    The slope of the tangent line is the derivative of thefunction, so
    f'(x)=y'=sqrt(x)*1'-1sqrt(x)'/sqrt(x)^2=-sqrt(x)/sqrt(x)^2,and install a for x, which doesn't mean anything until you get topart b, where you install the points (x,y) for their respectivevalues in the function.
  3. batur tekelio?lu
    batur tekelio?lu
    14 b
    = x^(-1/2)
    y' = -1/2*[x^(-1/3)......................1
    Put x= 1 in eqn 1,
    y ' = slope = -1/2
    Equation of tangent => (y-y1)=m(x-x1)
    Here m = -1/2 and x1,y1 = 1,1 Substitutethe valus and get the answers. Similarly you can do the otherone

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