Question

let f and g be polynomial functions, where f(x)=x^4 + 6x^2+8 and selected values of g are shown in the table below.

x	- 7	- 4	- 1	0	1	4	7
g(x)	3	5	7	0	7	5	3

(a) (i) find the values of x, as a decimal approximation, where f(x)=15, or indicate no such x - values exist.
(ii) find all the real zeroes of f or indicate that there are no real zeroes.
(b) (i) find the values of x, as a decimal approximation, where g(x)=5, or indicate no such x - values exist.
(ii) determine the end behavior of g(x) as x increases without bound. express your answer using the mathematical notation of a limit.
(c) (i) determine if g is an even function, an odd function, or neither.
(ii) give a reason for your answer based on the definition of even/odd functions and the table values of g(x).
write your responses to this question only on the designated pages in the separate free response booklet. write your solution to each part in the space provided for that part.

Explanation:

Step1: Solve \(f(x)=15\)

Set \(x^{4}+6x^{2}+8 = 15\). Let \(u = x^{2}\), then the equation becomes \(u^{2}+6u + 8-15=0\), i.e., \(u^{2}+6u - 7 = 0\).
Factor the quadratic equation: \((u + 7)(u - 1)=0\). So \(u=-7\) or \(u = 1\). Since \(u=x^{2}\geq0\), we discard \(u=-7\). When \(u = 1\), \(x^{2}=1\), so \(x=\pm1\).

Step2: Find real - zeroes of \(f(x)\)

Set \(x^{4}+6x^{2}+8 = 0\). Let \(u = x^{2}\), then \(u^{2}+6u + 8=0\). Factor: \((u + 2)(u + 4)=0\). So \(u=-2\) or \(u=-4\). But since \(u=x^{2}\geq0\) for real \(x\), there are no real zeroes of \(f(x)\).

Step3: Find \(x\) for \(g(x)=5\)

From the table, when \(g(x)=5\), \(x=-4\) or \(x = 4\).

Step4: Determine end - behavior of \(g(x)\)

Since \(g(x)\) is a polynomial function and the values of \(g(x)\) for \(x=-7\) and \(x = 7\) are the same (\(g(-7)=g(7)=3\)), and \(g(x)\) is symmetric about the \(y\) - axis (from the table values), as \(x\to+\infty\), \(g(x)\to3\) (because of the symmetry and the values at large \(|x|\)). In limit notation, \(\lim_{x\to+\infty}g(x)=3\).

Step5: Determine if \(g(x)\) is even or odd

A function \(y = g(x)\) is even if \(g(-x)=g(x)\) for all \(x\) in the domain, and odd if \(g(-x)=-g(x)\) for all \(x\) in the domain. From the table, \(g(-7)=g(7) = 3\), \(g(-4)=g(4)=5\), \(g(-1)=g(1)=7\). So \(g(x)\) is an even function. The reason is that for the values of \(x\) in the table, \(g(-x)=g(x)\) which satisfies the definition of an even function.

Answer:

(A) (i) \(x = 1\) and \(x=-1\)
(ii) No real zeroes
(B) (i) \(x=-4\) and \(x = 4\)
(ii) \(\lim_{x\to+\infty}g(x)=3\)
(C) (i) Even function
(ii) For the values of \(x\) in the table, \(g(-x)=g(x)\) which satisfies the definition of an even function.