3 Function/Pre-Calculus Questions

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• October 27th 2006, 02:05 PM
SportfreundeKeaneKent
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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• October 27th 2006, 02:27 PM
CaptainBlack
1. The roots of $x^3-ax^2-3kx+m=0$ are $2, -1, 3$ find the values of $a, k, m$.

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

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

expanding the brackets gives:

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

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

RonL
• October 27th 2006, 02:31 PM
topsquark
If $x^3-ax^2-3kx+m=0$ 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:
$(x - 2)(x - (-1))(x - 3) = (x -2)(x+1)(x-3)$

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

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

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

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

-Dan
• October 27th 2006, 02:38 PM
CaptainBlack
$
\frac{2x-1}{x-4}>1,\ \ x \ne 4
$

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

$
2x-1>x-4
$
,

or rearranging:

$
x>-3
$
,

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

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

$
2x-1$
,

or rearranging:

$
x<-3
$
.

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

RonL
• October 27th 2006, 02:40 PM
topsquark
Solve $\frac{2x-1}{x-4}>1$

First:

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

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

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

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 $(- \infty , -3) \cup (-3, 4) \cup (4, \infty)$ and test each interval:

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

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

-Dan

(Hi CaptainBlack! :) )
• October 27th 2006, 02:47 PM
CaptainBlack
For $f(x)=x^2+kx+k$ determine all values of $k$ such that $f(x)>0$.

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

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

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

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

RonL
• October 27th 2006, 02:52 PM
Soroban
Hello, SportfreundeKeaneKent!

Quote:

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

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

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

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

Quote:

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

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

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

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

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

Quote:

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

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

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

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

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

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

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

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

• October 28th 2006, 02:37 PM
SportfreundeKeaneKent
Terrific replies. I really appreciate it and I can do other problems like these now so thanks again.