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    Mathematics - Calculus

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    I really need help on these 4 problems and if you could please post your steps on how to do them, I'd greatly appreciate it.

    #1 Find an equation for the line tangent to the curve at the point defined by the given value of t.

    Also, find the value of d2y/dx2 at this point x=6t, y = √t, t = 1/25

    The equation for the line tangent to the curve at t=1/25 is?

    d2y/dx2| t= 1/25 = ?

    #2. Find the function's absolute maximum and minimum values on the given interval.

    F(x) = x 5/3, -8<x<1

    The absolute maximum of the function f(x) = x5/3 on the interval -8<x<1 has a value of ?

    The absolute minimum of the function f(x) = x5/3 on the interval -8<x<1 has a value of?

    #3. Find the absolute maximum and minimum values of the function on the given interval. F(x) = 1/3 x+5, -1 < x < 3

    The absolute maximum of the function f(x) = 1/3 x+5 on the interval -1<x<3 has a value of?

    The absolute minimum of the function (x)=1/3x+5 on the interval -1<x<3 has a value of ?

    #4 Find the value or values of c that satisfy the equation f(b)-f(a)/b-a = f(c) in the conclusion of the Mean Value Theorem for the following function and interval.

    F(x) = 5x(squared) + 2x-3 [-1,0]

    The values(s) of c that satisfy the equation f(b)-f(a)/b-a = f(c) is/are ?

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    Solution Preview

    The solution file is attached.

    Find an equation for the line tangent to the curve at the point defined by the given value of t.

    Also, find the value of d2y/dx2 at this point x=6t, y = √t, t = 1/25

    The equation for the line tangent to the curve at t=1/25 is?

    d2y/dx2| t= 1/25 = ?

    (a) dx/dt = 6, dy/dt = 1/2√t
    dy/dx = (dy/dt)/(dx/dt) = 1/(12√t) = 1/(12 √(1/25)) = 5/12
    When t = 1/25, x = 6t = 6/25 and y = √t = √(1/25) = 1/5
    The equation to the line ...

    Solution Summary

    A complete, neat and step-by-step solution to the calculus questions is provided in the attached file.

    $2.19

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