Let's analyze the given function [tex]\( f(x) = (5 - 6x)(2x - 4)^2 \)[/tex].
### (a) Finding the roots of [tex]\(f(x)\)[/tex]
To find the roots, we set the function equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ (5 - 6x)(2x - 4)^2 = 0 \][/tex]
This equation will be zero if either factor is zero.
1. [tex]\( 5 - 6x = 0 \)[/tex]
[tex]\[ 5 = 6x \][/tex]
[tex]\[ x = \frac{5}{6} \][/tex]
2. [tex]\( (2x - 4)^2 = 0 \)[/tex]
[tex]\[ 2x - 4 = 0 \][/tex]
[tex]\[ 2x = 4 \][/tex]
[tex]\[ x = 2 \][/tex]
So the roots of [tex]\( f(x) \)[/tex] are [tex]\( x = \frac{5}{6} \)[/tex] and [tex]\( x = 2 \)[/tex].
### (b) As [tex]\( x \rightarrow \infty, f(x) \rightarrow \)[/tex]
To understand the behavior of the function as [tex]\( x \)[/tex] approaches infinity, we need to look at the dominant terms in the function:
[tex]\[ f(x) = (5 - 6x)(2x - 4)^2 \][/tex]
As [tex]\( x \)[/tex] becomes very large, the term [tex]\( -6x \)[/tex] will dominate [tex]\( 5 \)[/tex] in the first factor, and the term [tex]\( 2x \)[/tex] will dominate [tex]\(-4 \)[/tex] in the second factor. This simplifies the function to approximately:
[tex]\[ f(x) \approx (-6x)(4x^2) \][/tex]
[tex]\[ f(x) \approx -24x^3 \][/tex]
As [tex]\( x \)[/tex] tends to infinity, [tex]\( x^3 \)[/tex] grows larger and larger, but since the coefficient is negative, [tex]\( -24x^3 \)[/tex] tends to negative infinity.
Therefore,
[tex]\[ \lim_{{x \to \infty}} f(x) = -\infty \][/tex]
### (c) As [tex]\( x \rightarrow -\infty, f(x) \rightarrow \)[/tex]
Similarly, to understand the behavior of the function as [tex]\( x \)[/tex] approaches negative infinity, we again consider the dominant terms:
[tex]\[ f(x) = (5 - 6x)(2x - 4)^2 \][/tex]
When [tex]\( x \)[/tex] is very large in the negative sense, the term [tex]\( -6x \)[/tex] will dominate [tex]\( 5 \)[/tex] and the term [tex]\( 2x \)[/tex] will dominate [tex]\(-4\)[/tex]. Simplifying the function, we get:
[tex]\[ f(x) \approx (-6x)(4x^2) \][/tex]
[tex]\[ f(x) \approx -24x^3 \][/tex]
As [tex]\( x \)[/tex] tends to negative infinity, [tex]\( x^3 \)[/tex] becomes very large in the negative sense, but since cubing a negative number results in a negative number, it will be a large negative number. Given the negative coefficient, [tex]\( -24x^3 \)[/tex] tends to positive infinity.
Therefore,
[tex]\[ \lim_{{x \to -\infty}} f(x) = \infty \][/tex]
To summarize:
(a) The roots of [tex]\( f(x) \)[/tex] are [tex]\( x = \frac{5}{6} \)[/tex] and [tex]\( x = 2 \)[/tex].
(b) As [tex]\( x \rightarrow \infty, f(x) \rightarrow -\infty \)[/tex]
(c) As [tex]\( x \rightarrow -\infty, f(x) \rightarrow \infty \)[/tex]
