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Math Help - Calculus

  1. #1
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    Calculus

    Hi. I was wondering if anyone could help me on these two problems. I'm having some difficulty with the problems, so if someone could please explain it, I'd be very grateful. Thanks.

    (multiple choice)
    1. Let f be the function given by f(x) = x^3 - 6x^2 + 7x + 3. The tangent line to the graph at x = 4 is used to approximate f(4.2). What is the error in this approximation?
    A) 0 B) 0.008 C) 0.248 D) 0.400 E) 0.648
    2. Let f be the function given by f(x) = (x^2)(lnx). For what value of x is the slope of the line tangent to the graph at (x, f(x)) equal to 2?
    A) 1.305 B) 1.548 C) 2.000 D) 2.548 E) 4.773
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  2. #2
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    Hello, turtle!

    #2 is a strange one . . .


    2. Given: .f(x) .= .xln(x)
    For what value of x is the slope of the line tangent to the graph equal to 2?

    . . A) 1.305 . . B) 1.548 . . C) 2.000 . . D) 2.548 . . E) 4.773

    The slope is: .f'(x) .= .x(1/x) + 2xln(x) .= .x + 2x.ln(x) .= .x[1 + ln(x)]


    When is f'(x) = 2?

    But there is no method for solving: .x[1 + ln(x)] .= .2

    So we must test the multiple-choice answers ... good ol' trial-and-error!


    We find that: .f(1.305) .= .1.305[1 + ln(1.305)] .= .1.999789936

    Therefore: .A) 1.305

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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by turtle View Post
    1. Let f be the function given by f(x) = x^3 - 6x^2 + 7x + 3. The tangent line to the graph at x = 4 is used to approximate f(4.2). What is the error in this approximation?
    A) 0 B) 0.008 C) 0.248 D) 0.400 E) 0.648
    f'(x) = 3x^2 - 12x + 7

    So at x = 4 the tangent line has a slope of f'(4) = 3*4^2 - 12*4 + 7 = 7

    Thus the tanget line is of the form
    y = 7x + b

    At x = 4, f(4) = -1 so the tangent line passes through the point (4, -1). Thus
    -1 = 7*4 + b ==> b = -29

    So finally, the tangent line is
    y = 7x - 29.

    The tangent line is the first approximation to the function value at x =4.2, so what is the y value of the tangent line at x = 4.2?
    y(4.2) = 7*4.2 - 29 = 0.4

    What is the actual value of the function at x = 4.2?
    f(4.2) = 4.2^3 - 6*4.2^2 + 7*4.2 + 3 = 0.648

    So the error in the estimate is 0.648 - 0.4 = 0.248 or answer C).

    -Dan
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