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Math Help - [SOLVED] Function Question....

  1. #1
    Member looi76's Avatar
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    [SOLVED] Function Question....

    Question:
    Express x^2 + 4x in the form (x + a)^2 + b , stating the numerical values of a and b.

    The functions f and g are defined as follows:


    f : x \mapsto x^2 + 4x,██ x \geq -2,

    g : x \mapsto x + 6,██  x \in R

    (i) Show that the equation gf(x) = 0 has no real roots.
    (ii) State the domain of f^{-1}, and find an expression in terms of x for f^{-1}(x).
    (iii) Sketch, in a single diagram, the graph of y = f(x) and y = f^{-1}(x), making clear the relationship between these graphs.


    Attempt:

    x^2 + 4x

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

    \Rightarrow (x + 2)^2 - 4

    a = 2 , b = -4

    (i) gf(x) = (x^2 + 4x) + 6
    gf(x) = (x + 2)^2 - 4 + 6
    gf(x) = (x + 2)^2 + 2

    (x + 2)(x + 2)
    \Rightarrow x^2 + 2x + 2x + 4
    \Rightarrow x^2 + 4x + 4

    gf(x) = x^2 + 4x + 6
    a = 1, b = 4, c = 6

    x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

    x = \frac{4 \pm \sqrt{4^2 - 4 \times 1 \times 6}}{2 \times 1}

    I can't complete because I get a negative value in the square root. Where did I go wrong?
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  2. #2
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    Quote Originally Posted by looi76 View Post
    Question:
    Express x^2 + 4x in the form (x + a)^2 + b , stating the numerical values of a and b.

    The functions f and g are defined as follows:

    f : x \mapsto x^2 + 4x,██ x \geq -2,

    g : x \mapsto x + 6,██  x \in R

    (i) Show that the equation gf(x) = 0 has no real roots.
    (ii) State the domain of f^{-1}, and find an expression in terms of x for f^{-1}(x).
    (iii) Sketch, in a single diagram, the graph of y = f(x) and y = f^{-1}(x), making clear the relationship between these graphs.

    Attempt:

    x^2 + 4x

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

    \Rightarrow (x + 2)^2 - 4

    a = 2 , b = -4

    (i) gf(x) = (x^2 + 4x) + 6
    gf(x) = (x + 2)^2 - 4 + 6
    gf(x) = (x + 2)^2 + 2

    (x + 2)(x + 2)
    \Rightarrow x^2 + 2x + 2x + 4
    \Rightarrow x^2 + 4x + 4

    gf(x) = x^2 + 4x + 6
    a = 1, b = 4, c = 6

    x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

    x = \frac{4 \pm \sqrt{4^2 - 4 \times 1 \times 6}}{2 \times 1}

    I can't complete because I get a negative value in the square root. Where did I go wrong?
    The fact that you get a negative value in the square root illustrates that the function x^2 + 4x + 6 has no real roots, which is what you want to show.
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  3. #3
    Member looi76's Avatar
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    Thanks icemanfan

    (ii) (y + 2)^2 - 4 = x
    (y + 2)^2 = x + 4
    y  + 2 = \sqrt{x  + 4}
    y = \sqrt{x + 4} - 2

    f^{-1}(x) = \sqrt{x + 4} - 2

    What is the domain and what is the range?
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  4. #4
    Member Danshader's Avatar
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    i hope you know the rules on real and imaginary roots:
    ax^2 + b.x + c

    if b^2 - 4.a.c > 0 you will have two different real roots
    if b^2 - 4.a.c = 0 you will have equal real roots
    if b^2 - 4.a.c < 0 you will have imaginary conjugate roots

    in your case:
    .
    .
    .
    = (x+2)^2 +2
    = x^2 + 4x +6 ====> a=1, b=4, c=6

    b^2 - 4.a.c = -8 (two imaginary roots)
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  5. #5
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    Quote Originally Posted by looi76 View Post
    Thanks icemanfan

    (ii) (y + 2)^2 - 4 = x
    (y + 2)^2 = x + 4
    y + 2 = \sqrt{x + 4}
    y = \sqrt{x + 4} - 2

    f^{-1}(x) = \sqrt{x + 4} - 2

    What is the domain and what is the range?
    This is a tricky question. It turns out that you can also find the inverse function y = -\sqrt{x + 4} - 2. But for the function
    y = \sqrt{x + 4} - 2, the domain is x \geq -4, all of the numbers that will maintain a nonnegative value within the square root. The range for this function is y \geq -2, since the square root is always greater than or equal to zero. See if you can find the domain and range for the other inverse function.
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  6. #6
    Member looi76's Avatar
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    Quote Originally Posted by Danshader View Post
    i hope you know the rules on real and imaginary roots:
    ax^2 + b.x + c

    if b^2 - 4.a.c > 0 you will have two different real roots
    if b^2 - 4.a.c = 0 you will have equal real roots
    if b^2 - 4.a.c < 0 you will have imaginary conjugate roots

    in your case:
    .
    .
    .
    = (x+2)^2 +2
    = x^2 + 4x +6 ====> a=1, b=4, c=6

    b^2 - 4.a.c = -8 (two imaginary roots)

    Thanks Danshader

    What does Discriminant mean? In the textbook it is written that "Discriminant is -8 < 0"
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  7. #7
    Member looi76's Avatar
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    Quote Originally Posted by looi76 View Post

    (iii) Sketch, in a single diagram, the graph of y = f(x) and y = f^{-1}(x), making clear the relationship between these graphs.


    Can someone please help me with the diagram? I can't draw it!
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  8. #8
    Member Danshader's Avatar
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    b^2 - 4a.c is called the discriminant.
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  9. #9
    Moo
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    Quote Originally Posted by looi76 View Post

    Thanks Danshader

    What does Discriminant mean? In the textbook it is written that "Discriminant is -8 < 0"
    Hello,

    Discriminant - Wikipedia, the free encyclopedia

    To sum up :

    If you have an equation ax^2+bx+c=0

    \Delta=b^2-4ac

    If \Delta <0, the equation has no real root and ax^2+bx+c is of the same sign as a.

    If \Delta=0, ax^2+bx+c=\left(x+\frac{b}{2a}\right)^2

    If \Delta>0, ax^2+bx+c=\left(x-\frac{-b+\sqrt{\Delta}}{2a}\right)\left(x-\frac{-b-\sqrt{\Delta}}{2a}\right)

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  10. #10
    Member looi76's Avatar
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    Quote Originally Posted by Danshader View Post
    b^2 - 4a.c is called the discriminant.
    Quote Originally Posted by Moo View Post
    Hello,

    Discriminant - Wikipedia, the free encyclopedia

    To sum up :

    If you have an equation ax^2+bx+c=0

    \Delta=b^2-4ac

    If \Delta <0, the equation has no real root and ax^2+bx+c is of the same sign as a.

    If \Delta=0, ax^2+bx+c=\left(x+\frac{b}{2a}\right)^2

    If \Delta>0, ax^2+bx+c=\left(x-\frac{-b+\sqrt{\Delta}}{2a}\right)\left(x-\frac{-b-\sqrt{\Delta}}{2a}\right)

    Thanks Danshader and Moo. Can you guys please help me with the graph?
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