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Math Help - A few questions

  1. #1
    Junior Member NAPA55's Avatar
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    A few questions

    Could someone please help me with these couple questions? Instructions on how to do the first two would be great, but for 3 and 4 I know how to do them but I'm not getting the right answers.

    1. Determine a quadratic function f(x)=ax+bx+c whose graphy passes through the point (2,19) and that has a horizontal tangent at (-1,-8).

    2. For what values of x do the curves y=(1+x) and y=2x6 (6 is an exponent on the x) have the same slope?

    3. Differentiate y= (1+√x / x2/3 ), where 2/3 is an exponent on that x in the denominator.

    4. Differentiate y= (√(1-x) / 1-x ).

    Thanks a ton!!!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NAPA55 View Post
    1. Determine a quadratic function f(x)=ax+bx+c whose graphy passes through the point (2,19) and that has a horizontal tangent at (-1,-8).
    The function passes through (2, 19), so we know that
    19 = 4a + 2b + c

    And the function passes through (-1, -8), so
    -8 = a - b + c

    The derivative of the function is
    f^{\prime} = 2ax + b

    We know that the tangent is horizontal (that the first derivative is equal to 0) at x = -1:
    0 = -2a + b

    Three equations, three unknowns.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NAPA55 View Post
    2. For what values of x do the curves y=(1+x) and y=2x6 (6 is an exponent on the x) have the same slope?
    The equations have the same slope, so the derivatives are equal for those x values. So find the derivatives, set them equal to each other, and solve for x.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NAPA55 View Post
    3. Differentiate y= (1+√x / x2/3 ), where 2/3 is an exponent on that x in the denominator.
    y = \left ( \frac{1 + \sqrt{x}}{x^{2/3}} \right ) ^3

    This is an exercise in the chain rule:
    y^{\prime} = \left [ 3 \left ( \frac{1 + \sqrt{x}}{x^{2/3}} \right )^2 \right ] \cdot \left [ \frac{\frac{1}{2 \sqrt{x}} \cdot x^{2/3} - (1 + \sqrt{x} ) \cdot \frac{2}{3}x^{-1/3} }{x^{4/3}} \right ]

    The first set of [ ] is the derivative of the "outer" function, the cube. The second set of [ ] is the derivative of the fraction inside the ( ). Of course, you still need to simplify this.

    -Dan
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  5. #5
    Junior Member NAPA55's Avatar
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    Quote Originally Posted by topsquark View Post
    y = \left ( \frac{1 + \sqrt{x}}{x^{2/3}} \right ) ^3

    This is an exercise in the chain rule:
    y^{\prime} = \left [ 3 \left ( \frac{1 + \sqrt{x}}{x^{2/3}} \right )^2 \right ] \cdot \left [ \frac{\frac{1}{2 \sqrt{x}} \cdot x^{2/3} - (1 + \sqrt{x} ) \cdot \frac{2}{3}x^{-1/3} }{x^{4/3}} \right ]

    The first set of [ ] is the derivative of the "outer" function, the cube. The second set of [ ] is the derivative of the fraction inside the ( ). Of course, you still need to simplify this.

    -Dan
    That's what I had so far, but with all the fractions I got super confused on the simplification.

    And for #1, what do I do with the three equations and the three unknowns?

    Thanks for your help!!
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  6. #6
    Junior Member NAPA55's Avatar
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    Quote Originally Posted by topsquark View Post
    So find the derivatives, set them equal to each other, and solve for x.
    So wouldn't it be

    6x + 6x5 = 12x5
    6x = 6x5
    x=x5

    Now what?
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NAPA55 View Post
    So wouldn't it be

    6x + 6x5 = 12x5
    6x = 6x5
    x=x5

    Now what?
    x^5 = x^2

    x^5 - x^2 = 0

    x^2(x^3 - 1) = 0

    x^2(x - 1)(x^2 + x + 1) = 0

    Set each of these factors equal to 0 and you find that
    x = 0, 1
    (The third term has only complex zeros, so we may ignore the solutions from this factor.)

    -Dan
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