# 3 Function/Pre-Calculus Questions

• Oct 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.

http://img106.imageshack.us/my.php?i...ntitledju7.png

If that picture doesn't work, then this should:

http://img106.imageshack.us/img106/7...tledju7.th.png

[IMG]http://img106.imageshack.us/img106/7...tledju7.th.png[/IMG]
• Oct 27th 2006, 02:27 PM
CaptainBlack
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
• Oct 27th 2006, 02:31 PM
topsquark
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
• Oct 27th 2006, 02:38 PM
CaptainBlack
$\displaystyle \frac{2x-1}{x-4}>1,\ \ x \ne 4$

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
• Oct 27th 2006, 02:40 PM
topsquark
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

(Hi CaptainBlack! :) )
• Oct 27th 2006, 02:47 PM
CaptainBlack
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
• Oct 27th 2006, 02:52 PM
Soroban
Hello, SportfreundeKeaneKent!

Quote:

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

Quote:

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

Quote:

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

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