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Math Help - 2 problem exam questions

  1. #1
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    2 problem exam questions

    Hey i need to complete the following exam questions for my tutor for tomorrow but i am having trouble understanding them:

    1. Show that x^2+x+1>0 for all values of x. (3 marks)

    For questions such as these am i supposed to use proof or what... im just not sure how to approach such a simple equation.

    The 2nd question is a harder question involving calculus:

    2. The curve C has the equation y=\frac{x}{9+x^2}

    Find the coordinates of the turning points of C. (6 marks)

    I have got so far on the 2nd question:

    y=\frac{x}{9+x^2}

    Using quotient rule:

    \frac{d}{dx}(\frac{x}{9+x^2})=\frac{1(9+x^2)-x(2x)}{(9+x^2)^2}

    =\frac{9+x^2-2x^2}{(9+x^2)^2}

    =\frac{9-x^2}{(9+x^2)^2}

    In order to complete the question i must know some values of x or y to substitute in order to find the turning points, but i am not sure how i can get them with such little information.

    I would be grateful if someone could show me how to complete the question with full workings.

    Thanks in advance to anyone who helps out.
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  2. #2
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    Quote Originally Posted by Big_Joe View Post
    Hey i need to complete the following exam questions for my tutor for tomorrow but i am having trouble understanding them:

    1. Show that x^2+x+1>0 for all values of x. (3 marks)

    For questions such as these am i supposed to use proof or what... im just not sure how to approach such a simple equation.

    The 2nd question is a harder question involving calculus:

    2. The curve C has the equation y=\frac{x}{9+x^2}

    Find the coordinates of the turning points of C. (6 marks)

    I have got so far on the 2nd question:

    y=\frac{x}{9+x^2}

    Using quotient rule:

    \frac{d}{dx}(\frac{x}{9+x^2})=\frac{1(9+x^2)-x(2x)}{(9+x^2)^2}

    =\frac{9+x^2-2x^2}{(9+x^2)^2}

    =\frac{9-x^2}{(9+x^2)^2}

    In order to complete the question i must know some values of x or y to substitute in order to find the turning points, but i am not sure how i can get them with such little information.

    I would be grateful if someone could show me how to complete the question with full workings.

    Thanks in advance to anyone who helps out.
    For the first part, set the expression equal to zero and solve for x (you'll have to use the quadratic formula). You'll find that the discriminant is less than 0, and hence there are no real roots. Which means that the function does NOT cross the x axis into negative values of y. You can prove that it is always in positive values by stating that it doesn't cross the y axis, is continuous, and has at least one value in positive y.

    For the 2nd question, remember that to find the turning point, the first differential must be set to zero and then solve for x to get the x values of the turning point. The y values can then be determined from the original equation. Your first differential is a quotient so remember that only the numerator of a quotient can be equal to zero, the denominator equal to zero gives an undefined expression.
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  3. #3
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    Complete the square

    Hello Big_Joe
    Quote Originally Posted by Big_Joe View Post
    1. Show that x^2+x+1>0 for all values of x. (3 marks)
    1 You need to complete the square, as follows:

    Step 1: Make the coefficient of
    x^2 equal to 1 by taking out a factor, if necessary. Already 1, so not necessary.

    Step 2: Write inside a bracket
    x plus half coefficient of x from previous line. Write a power 2 outside the bracket:

    x^2+x+1 = \left(x + \frac{1}{2}\right)^2 + ...

    Step 3: Subtract the square of the number you just wrote inside the bracket; that was
    \frac{1}{2}. So:

    \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} ...

    Step 4: Replace the original constant, and simplify:

    \left(x + \frac{1}{2}\right)^2 - \frac{1}{4}+1

    =\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}

    Now look at what you've got. Whatever the value of x, the value of \left(x + \frac{1}{2}\right)^2 is never negative, because it's the square of a real number. So the minimum value of \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} is ... ? Can you see it now?

    2. The curve C has the equation y=\frac{x}{9+x^2}

    Find the coordinates of the turning points of C. (6 marks)

    I have got so far on the 2nd question:

    y=\frac{x}{9+x^2}

    Using quotient rule:

    \frac{d}{dx}(\frac{x}{9+x^2})=\frac{1(9+x^2)-x(2x)}{(9+x^2)^2}

    =\frac{9+x^2-2x^2}{(9+x^2)^2}

    =\frac{9-x^2}{(9+x^2)^2}
    Your working is correct, and you're very close. The turning points are where
    \frac{dy}{dx}=0, and that means:

    \frac{9-x^2}{(9+x^2)^2}=0

    Multiply both side by
    (9+x^2)^2:

    \implies 9-x^2=0

    Can you see what to do now?

    Grandad
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  4. #4
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    Ok from what you've said i can continue on the 2nd question by saying:

    9-x^2=0
    so x=\sqrt9
    so x=3

    From this: y=\frac{3}{9+3^2}
    So y=\frac{3}{18}

    So y=\frac{1}{6}

    OK i understand this so far, but how do i find the turning points. I did think i needed to find a 2nd derivative or is there another way?

    (P.S. Have finished editing now)
    Last edited by Big_Joe; December 29th 2008 at 06:10 AM.
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  5. #5
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    Personally, I do not like the term 'turning points'. That does not sound

    very mathematical. If they wanted critical points or stationary points, they

    should say so.

    If they want inflection points or points of concavity change, they should say so.

    But that's me.
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  6. #6
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    Quote Originally Posted by galactus View Post
    Personally, I do not like the term 'turning points'. That does not sound

    very mathematical. If they wanted critical points or stationary points, they

    should say so.

    If they want inflection points or points of concavity change, they should say so.

    But that's me.
    Assuming they are looking for the min/max points (which i think is more likely, however it is annoying that they don't specify) how would i continue to solve this?
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  7. #7
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    Turning points

    Hello Big_Joe
    Quote Originally Posted by Big_Joe View Post
    Assuming they are looking for the min/max points (which i think is more likely, however it is annoying that they don't specify) how would i continue to solve this?
    Possibly 'turning points' is a specifically British term, but that is definitely what min/max points are sometimes called in the UK.

    So... to resume your question.

    The second commandment of mathematics is When thou takest a square root, forget not thy plus or minus sign. (The first commandment is Thou shalt not divide by zero.) So, don't forget that \sqrt{9} has two values!

    Next, yes you do need to differentiate again. But don't be afraid of it. Just use the quotient rule again - you've done it successfully once - and then put in the two values of x to determine the sign of the result. You don't need to work out the actual value of
    \frac{d^2y}{dx^2} - just its sign. And the denominator of the fraction will be (something) to the power 4, which is bound to be positive, so you can more or less ignore that!

    Have a go.

    Grandad
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  8. #8
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    Quote Originally Posted by Big_Joe View Post
    Assuming they are looking for the min/max points (which i think is more likely, however it is annoying that they don't specify) how would i continue to solve this?
    I have to disagree with galacticus here: "turning point" is a mathematical term and if the question had asked for "critical points" it would have been asking a completely different question. A turning point is always a critical point but a critical point may not be a turning point: a critical point is a point at which the derivative is 0 (or does not exist) while a turning point is a point where the derivative changes sign. The deriviative of a function is not necessarily continuous but still must satisfy the "intermediate value property"- at a point where the derivative changes sign, it must be 0. "Turning point" is a more specific term and so is completely correct in this problem.

    You can find any critical point (and so any turning point) by setting the derivative equal to 0. You will need to check if it is a turning point, perhaps by checking that the second derivative is not 0 (which must happen if a point is a critical point but not a turning point).

    For example, if y= 3x^4- 4x^3, y'= 12x^3- 12x^2[/itex] which is 0 at x= 0 and x= 1, the critical points. But y"= 36x^2- 24x which is 0 at a x= 0 but not at x= 1. x= 0 is a turning point but x= 1 is not.
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  9. #9
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    Quote Originally Posted by Grandad View Post
    Hello Big_Joe

    Possibly 'turning points' is a specifically British term, but that is definitely what min/max points are sometimes called in the UK.

    So... to resume your question.

    The second commandment of mathematics is When thou takest a square root, forget not thy plus or minus sign. (The first commandment is Thou shalt not divide by zero.) So, don't forget that \sqrt{9} has two values!

    Surely you didn't mean to say that! Yes, the equation x^2= 9 has two roots: 3 and -3. But only one of those is equal to [itex]\sqrt{9}[/tex]. That's why we must write "-" before \sqrt{9} in order to show the other root to the equation!

    Next, yes you do need to differentiate again. But don't be afraid of it. Just use the quotient rule again - you've done it successfully once - and then put in the two values of x to determine the sign of the result. You don't need to work out the actual value of
    \frac{d^2y}{dx^2} - just its sign. And the denominator of the fraction will be (something) to the power 4, which is bound to be positive, so you can more or less ignore that!

    Have a go.

    Grandad
    Last edited by mr fantastic; December 30th 2008 at 02:23 PM.
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  10. #10
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    I must plead ignorance here. I have not heard that term before in regards to

    critical points. To make a guess, I would assume(incorrectly) it meant change

    of concavity. Now I know better and learned something.
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  11. #11
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    Thanks for trying to help everyone this is how i continued. Have i got it correct?
    So \sqrt9=3 or -3 so y=\frac{1}{6} or -\frac{1}{6}

    2nd derivative:

    \frac{d^2y}{dx^2}=\frac{-2x(9+x^2)^2-2(9+x^2)(2x)(9-x^2)}{((9+x^2)^2)^2}

    =\frac{-2x(9+x^2)^2-4x(9+x^2)(9-x^2)}{(9+x^2)^4}

    =\frac{-2x(9+x^2)-4x(9-x^2)}{(9+x^2)^3}

    =\frac{-18x-2x^3-36x+4x^3}{(9+x^2)^3}

    =\frac{2x^3-54x}{(9+x^2)^3}

    f"(3)=\frac{2(3^3)-54(3)}{(9+3^2)^3}

    =\frac{53-162}{18^3}

    =-\frac{108}{5883}

    f"(-3)=\frac{108}{5883}

    Oh and from this info what are actually the coordinates of the turning points?
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  12. #12
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    Max and min

    Hello Big_Joe

    Looks perfect to me. So now you know that:

    When x = 3, y = \frac{1}{6}, \frac{dy}{dx}=0 and \frac{d^2y}{dx^2} <0 \implies \frac{dy}{dx} is decreasing at this point \implies we have a
    maximum turning point at (3, 1/6).

    And when x = -3, ... you can complete it now.

    Grandad

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  13. #13
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    Thanks i got this sorted now!
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