Question

1. given ( f(x)=x^{3}-7 x^{2}+12 x ).

a. determine whether the function is even, odd, or neither using the algebraic test.
b. use limits to describe the end - behavior of the function.
c. find all real zeros.
d. for what interval(s) is ( f(x)>0 )?

Explanation:

A. Determine whether the function is even, odd, or neither using the algebraic test.

Step1: Find \(f(-x)\)

Given \(f(x)=x^{3}-7x^{2}+12x\), then \(f(-x)=(-x)^{3}-7(-x)^{2}+12(-x)=-x^{3}-7x^{2}-12x\)

Step2: Check for even function (\(f(-x)=f(x)\))

\(f(x)=x^{3}-7x^{2}+12x\) and \(f(-x)=-x^{3}-7x^{2}-12x\). Since \(f(-x)
eq f(x)\), it is not even.

Step3: Check for odd function (\(f(-x)=-f(x)\))

\(-f(x)=-(x^{3}-7x^{2}+12x)=-x^{3}+7x^{2}-12x\). Since \(f(-x)
eq -f(x)\), it is not odd.

Step1: Analyze the leading term

The function \(y = f(x)=x^{3}-7x^{2}+12x\) is a polynomial function. The leading term is \(x^{3}\) (degree \(n = 3\), coefficient \(a=1\))

Step2: Find \(\lim_{x

ightarrow\infty}f(x)\)
\(\lim_{x
ightarrow\infty}(x^{3}-7x^{2}+12x)=\lim_{x
ightarrow\infty}x^{3}(1 - \frac{7}{x}+\frac{12}{x^{2}})\)
As \(x
ightarrow\infty\), \(\frac{7}{x}
ightarrow0\) and \(\frac{12}{x^{2}}
ightarrow0\). So \(\lim_{x
ightarrow\infty}(x^{3}-7x^{2}+12x)=\infty\)

Step3: Find \(\lim_{x

ightarrow-\infty}f(x)\)
\(\lim_{x
ightarrow-\infty}(x^{3}-7x^{2}+12x)=\lim_{x
ightarrow-\infty}x^{3}(1 - \frac{7}{x}+\frac{12}{x^{2}})\)
As \(x
ightarrow-\infty\), \(\frac{7}{x}
ightarrow0\) and \(\frac{12}{x^{2}}
ightarrow0\). So \(\lim_{x
ightarrow-\infty}(x^{3}-7x^{2}+12x)=-\infty\)

Step1: Factor the function

\(f(x)=x^{3}-7x^{2}+12x=x(x^{2}-7x + 12)\)

Step2: Factor the quadratic

\(x^{2}-7x + 12=(x - 3)(x - 4)\)
So \(f(x)=x(x - 3)(x - 4)\)

Step3: Set \(f(x)=0\)

\(x(x - 3)(x - 4)=0\)
Using the zero - product property: \(x=0\) or \(x - 3=0\) or \(x - 4=0\)

Answer:

The function \(f(x)=x^{3}-7x^{2}+12x\) is neither even nor odd.

B. Use limits to describe the end - behavior of the function.