Answer:
(a) Domain: (-∞ ,∞) Range: (-∞ ,∞)
[tex]\textsf{As} \; x \rightarrow -\infty, \; f(x) \rightarrow - \infty[/tex]
[tex]\textsf{As} \; x \rightarrow +\infty, \; f(x) \rightarrow + \infty[/tex]
(b) x = 6
(c) See below.
(d) g(x) = (x - 6)(x² - 4x + 1)
(e) x = 6, x = 2 + √3, x = 2 - √3
Step-by-step explanation:
Given polynomial:
[tex]g(x)=x^3-10x^2+25x-6[/tex]
Part (a)
The domain of a function is the set of all possible input values (x-values).
The range of a function is the set of all possible output values (y-values).
Both the domain and range are unrestricted, therefore:
- Domain: (-∞ ,∞)
- Range: (-∞ ,∞)
As the leading coefficient is positive and the degree of the polynomial is odd, the end behavior of the function is:
- [tex]\textsf{As} \; x \rightarrow -\infty, \; f(x) \rightarrow - \infty[/tex]
- [tex]\textsf{As} \; x \rightarrow +\infty, \; f(x) \rightarrow + \infty[/tex]
Part (b)
Rational Root Theorem
[tex]\dfrac{p}{q}=\dfrac{\textsf{a factor of the last term $(a_0)$}}{\textsf{a factor of the first term $(a_n)$}}[/tex]
where:
[tex]f(x)=a_n x^n+a_{n-1} x^{n-1} +... + a_2 x^2+a_1 x +a_0[/tex]
Identify the factors (both positive and negative) of the constant of the polynomial. The factors are the possible values of p:
Identify the factors (both positive and negative) of the leading coefficient of the polynomial. These factors are the possible values of q.
Find each possible value of p/q
[tex]\implies \dfrac{p}{q}=\dfrac{\pm 1}{\pm 1}, \dfrac{\pm 2}{\pm 1}, \dfrac{\pm 3}{\pm 1}, \dfrac{\pm 6}{\pm 1}=\pm1, \pm2, \pm 3, \pm6[/tex]
Therefore, the possible rational roots of g(x) are:
Substitute each of the possible roots into g(x). Any root that results in f(x) = 0 is an actual rational root.
[tex]x=-1 \implies g(-1)=(-1)^3-10(-1)^2+25(-1)-6=-42[/tex]
[tex]x=1 \implies g(1)=(1)^3-10(1)^2+25(1)-6=10[/tex]
[tex]x=-2 \implies g(-2)=(-2)^3-10(-2)^2+25(-2)-6=-104[/tex]
[tex]x=2 \implies g(2)=(2)^3-10(2)^2+25(2)-6=12[/tex]
[tex]x=-3 \implies g(-3)=(-3)^3-10(-3)^2+25(-3)-6=-198[/tex]
[tex]x=3 \implies g(3)=(3)^3-10(3)^2+25(3)-6=6[/tex]
[tex]x=-6 \implies g(-6)=(-6)^3-10(-6)^2+25(-6)-6=-732[/tex]
[tex]x=6 \implies g(6)=(6)^3-10(6)^2+25(6)-6=0[/tex]
Therefore, the actual rational root is x = 6.
Part (c)
Descartes' Rule of Signs tells us the maximum number of positive and negative roots.
Positive root case
[tex]g(x)=+x^3-10x^2+25x-6[/tex]
As there are 3 sign changes, the maximum possible number of positive roots is 3.
Negative root case
[tex]\begin{aligned}g(-x)&=(-x)^3-10(-x)^2+25(-x)-6\\ &=-x^3-10x^2-25x-6 \end{aligned}[/tex]
As there are no sign changes, there are no negative roots.
Part (d)
Remainder Theorem
When we divide a polynomial f(x) by (x − c) the remainder is f(c).
Factor Theorem
If f(x) is a polynomial, and f(a) = 0, then (x – a) is a factor of f(x).
From part (b) we know that f(6) = 0, so (x - 6) is a factor of g(x):
Use long division to find the other factor:
[tex]\large \begin{array}{r}x^2-4x+1\phantom{)}\\x-6{\overline{\smash{\big)}\,x^3-10x^2+25x-6\phantom{)}}}\\{-~\phantom{(}\underline{(x^3-6x^2)\phantom{-b000000)}}\\-4x^2+25x-6\phantom{)}\\-~\phantom{()}\underline{(-4x^2+24x)\phantom{))..)}}\\x-6\phantom{)}\\-~\phantom{()}\underline{(x-6)\phantom{}}\\0 \phantom{)}\end{array}[/tex]
(x² - 4x + 1) cannot be factored further.
Therefore the factored polynomial is:
- [tex]g(x)=(x-6)(x^2-4x+1)[/tex]
Part (e)
The zeros of a polynomial are the values of x when f(x) = 0:
[tex]\implies g(x)=0[/tex]
[tex]\implies (x-6)(x^2-4x+1)=0[/tex]
Therefore:
[tex]\implies x-6=0[/tex]
[tex]\implies x^2-4x+1=0[/tex]
We have already established that x = 6 is a zero.
To find the other zeros, use the quadratic formula.
[tex]\boxed{\begin{minipage}{3.6 cm}\underline{Quadratic Formula}\\\\$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$\\\\when $ax^2+bx+c=0$ \\\end{minipage}}[/tex]
[tex]x^2-4x+1 \implies a=1, \;\;b=-4, \;\;c=1[/tex]
Therefore:
[tex]\implies x=\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(1)}}{2(1)}[/tex]
[tex]\implies x=\dfrac{4 \pm 2\sqrt{3}}{2}[/tex]
[tex]\implies x=2 \pm \sqrt{3}[/tex]
Therefore, the zeros of the given polynomial are:
- x = 6
- x = 2 + √3
- x = 2 - √3
Note: There are no complex zeros.
