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Thread: 3 Function/Pre-Calculus Questions

  1. #1
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    3 Function/Pre-Calculus Questions

    I can't solve these 3 solutions. Could someone post up solutions or tell me how to do them. Greatly appreciated.





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  2. #2
    Grand Panjandrum
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    1. The roots of $\displaystyle x^3-ax^2-3kx+m=0$ are $\displaystyle 2, -1, 3$ find the values of $\displaystyle a, k, m$.

    That the roots are $\displaystyle 2,\ -1,\ 3$ means that:

    $\displaystyle
    x^3-ax^2-3kx+m=(x-2)(x+1)(x-3)=0
    $

    expanding the brackets gives:

    $\displaystyle
    x^3-ax^2-3kx+m=x^3 - 4x^2 + x + 6=0
    $

    so $\displaystyle a=4,\ k=-1/3,\ m=6$

    RonL
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  3. #3
    Forum Admin topsquark's Avatar
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    If $\displaystyle x^3-ax^2-3kx+m=0$ has the roots x = 2, -1, 3. Find a, k, m.

    The simplest way to do this is to construct the polynomial that has those roots. So:
    $\displaystyle (x - 2)(x - (-1))(x - 3) = (x -2)(x+1)(x-3)$

    = $\displaystyle (x^2 - x - 2)(x - 3) = x^3 - x^2 - 2x - 3x^2 + 3x + 6$

    = $\displaystyle x^3 - 4x^2 + x + 6$

    So I get that
    -a = -4
    -3k = 1
    m = 6

    So a = 4, k = -1/3, m = 6.

    -Dan
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  4. #4
    Grand Panjandrum
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    $\displaystyle
    \frac{2x-1}{x-4}>1,\ \ x \ne 4
    $

    Suppose $\displaystyle x>4$ so $\displaystyle x-4>0$, now multiply through by $\displaystyle x-4$ to get:

    $\displaystyle
    2x-1>x-4
    $,

    or rearranging:

    $\displaystyle
    x>-3
    $,

    which tells us nothing extra, as we had assumed that $\displaystyle x-4>0$, or $\displaystyle x>4$.

    Instead suppose $\displaystyle x-4<0$, then:

    $\displaystyle
    2x-1<x-4
    $,

    or rearranging:

    $\displaystyle
    x<-3
    $.

    So our solution is $\displaystyle x>4$ or $\displaystyle x<-3$.

    RonL
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    Forum Admin topsquark's Avatar
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    Solve $\displaystyle \frac{2x-1}{x-4}>1$

    First:

    $\displaystyle \frac{2x-1}{x-4}-1>0$

    $\displaystyle \frac{2x-1}{x-4} - \frac{x-4}{x-4} > 0$

    $\displaystyle \frac{x+3}{x-4}>0$

    There are critical points where the numerator is 0 and where the denominator is 0. So we have critical points at:
    x + 3 = 0 --> x = -3
    x - 4 = 0 --> x = 4

    So break the real line up into $\displaystyle (- \infty , -3) \cup (-3, 4) \cup (4, \infty) $ and test each interval:

    $\displaystyle (-\infty, -3)$: $\displaystyle \frac{x+3}{x-4}>0$ (Check!)
    $\displaystyle (-3, 4)$: $\displaystyle \frac{x+3}{x-4}<0$ (Nope!)
    $\displaystyle (4, \infty)$: $\displaystyle \frac{x+3}{x-4}>0$ (Check!)

    So the solution set is $\displaystyle x \in (-\infty, -3) \cup (4, \infty)$

    -Dan

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    Grand Panjandrum
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    For $\displaystyle f(x)=x^2+kx+k$ determine all values of $\displaystyle k$ such that $\displaystyle f(x)>0$.

    Now for large $\displaystyle |x|\ f(x)>0$ as the leading term is dominant. So the question is realy asking us to find the values of k such that $\displaystyle f(x)=0$ has no real roots.

    From the quadratic formula we know the roots of $\displaystyle f(x)=0$ are:

    $\displaystyle
    x=\frac{-k \pm \sqrt{k^2-4k}}{2}
    $,

    and these are not real if $\displaystyle k^2-4k<0$, which is equivalent to $\displaystyle k<4$ and $\displaystyle k>0$.

    RonL
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  7. #7
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    Hello, SportfreundeKeaneKent!

    If the roots of the equation $\displaystyle x^3 - ax^2 - 3kx + m\:=\:0$ are $\displaystyle 2,\:-1$, and $\displaystyle 3$,
    find the values of $\displaystyle a,\:k$, and $\displaystyle m.$

    If $\displaystyle 2,\:-1$, and $\displaystyle 3$ are roots of the cubic,
    . . then $\displaystyle (x - 2),\:(x + 1),\;(x - 3)$ are factors of the cubic.

    Hence, the cubic is: .$\displaystyle (x - 2)(x + 1)(x - 3)\:=\:x^3- 4x^2+x - 6$

    Therefore: .$\displaystyle \begin{Bmatrix}\text{-}a = \text{-}4 \\ \text{-}3k = 1 \\ m = 6\end{Bmatrix}\;\Rightarrow\;\begin{Bmatrix}a = 4\\ k = \text{-}\frac{1}{3} \\ m = 6\end{Bmatrix} $


    Solve: .$\displaystyle \frac{2x-1}{x-4}\:>\:1$ . where $\displaystyle x \neq 4.$

    Multiply both sides by $\displaystyle (x - 4)$.

    If $\displaystyle x - 4 \,>\,0$ or $\displaystyle x \,<\,4$,
    . . we have: .$\displaystyle 2x - 1 \,>\,x - 4\quad\Rightarrow\quad x > -3$
    The "stronger" inequality is: .$\displaystyle x > 4$

    If $\displaystyle x - 4 \,<\,0$ or $\displaystyle x \,<\,4$,
    . . we have: .$\displaystyle 2x - 1 \,< \,x - 4\quad\Rightarrow\quad x < -3$
    The "stronger" inequality is: .$\displaystyle x < -3$

    Solution: .$\displaystyle \boxed{x < -3\;\text{ or }\;x > 4}$



    For $\displaystyle f(x)\:=\:x^2 + kx + k$, determine all the values of $\displaystyle k$ such that $\displaystyle f(x) > 0.$

    If $\displaystyle f(x) > 0$, the parabola is completely above the x-axis.

    If it had x-intercepts, they would occur where $\displaystyle f(x) = 0$.
    . . That is: .$\displaystyle x^2 + kx + k\:=\:0$

    Quadratic Formula: .$\displaystyle x \:=\:\frac{-k \pm \sqrt{k^2 - 4k}}{2}$

    To have no x-intercepts, the discriminant must be negative.
    . . That is: .$\displaystyle k^2 - 4k \:< \:0$

    Complete the square: .$\displaystyle k^2 - 4k + 4 \:< \:4$

    And we have: .$\displaystyle (k - 2)^2 \:< \:4\quad\Rightarrow\quad |k-2|\:<\:2$

    Hence: .$\displaystyle -2 \:<\:k-2 \:<\:2\quad\Rightarrow\quad \boxed{0 \:< \:k\:<\:4}$

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  8. #8
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    Terrific replies. I really appreciate it and I can do other problems like these now so thanks again.
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