Calculus exam review, PLEASE HELP!!

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for each function, determine the equation of the tangent line to the graph for the indicated value of the variable.

a) y = 3cosx + 4sinx + 1 , when x = pie

b) y = x - 3cosx , when x = (5pie)/3

  • Calculus exam review, PLEASE HELP!! -

    π is PI not PIE!

    a)
    y'(x) = -3sinx + 4cosx
    y'(π) = -3sin π + 4cos π = -4

    y(π) = 3(-1) + 0 + 1 = -2
    thus, the line through (π,-2) having slope -4 is

    (y+2) = -4(x-pi)

    b) y' = 1 + 3sinx
    y'(5π/3) = 1 + 3(-√3/2) = 1 - 3√3/2
    y(5π/3) = 5π/3 - 3(1/2) = 5π/3 - 3/2

    (y-(5π/3 - 3/2) = (1 - 3√3/2)(x-5π/3)

  • Calculus exam review, PLEASE HELP!! -

    thank you so much steve!

    for b) (y-(5π/3 - 3/2) = (1 - 3√3/2)(x-5π/3)

    is the square root for the whole 3/2 or is it just root 3 OVER 2.

  • Calculus exam review, PLEASE HELP!! -

    I think by now you know the value of cos(pi/3)

  • Calculus exam review, PLEASE HELP!! -

    (y-(5π/3 - 3/2) = (1 - (3√3)/2)(x-5π/3)

    like that ? :(

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