Question

on a separate sheet of paper, graph (f(x)=x^{4}-2x^{2}-2). then, estimate the x - coordinates at which each zero and relative extrema occur, and state the domain and range. zeros between (x=) blank and (x = - 1), and between (x = 1) and (x=) blank. relative maximum at (x=) blank. relative minimum at (x=) blank and (x = 1). (d=) all real numbers, (r={f(x)|f(x)geq) blank . need help with this question? get a hint.

Explanation:

Step1: Find the zeros

Set $y = x^{4}-2x^{2}-2$. Let $t = x^{2}$, then the equation becomes $t^{2}-2t - 2=0$. Using the quadratic formula $t=\frac{2\pm\sqrt{4 + 8}}{2}=\frac{2\pm2\sqrt{3}}{2}=1\pm\sqrt{3}$. Since $t = x^{2}\geq0$, $t=1 + \sqrt{3}$ (because $1-\sqrt{3}<0$), and $x=\pm\sqrt{1+\sqrt{3}}\approx\pm1.65$. The zeros are between $x=- 2$ and $x=-1$, and between $x = 1$ and $x = 2$.

Step2: Find the derivative

Differentiate $y=x^{4}-2x^{2}-2$ with respect to $x$. $y'=4x^{3}-4x=4x(x^{2}-1)=4x(x - 1)(x + 1)$.

Step3: Find the relative extrema

Set $y'=0$, we get $x=-1,0,1$. Using the second - derivative test, $y'' = 12x^{2}-4$. When $x = 0$, $y''=-4<0$, so there is a relative maximum at $x = 0$. When $x=\pm1$, $y''=12 - 4 = 8>0$, so there are relative minima at $x=-1$ and $x = 1$.

Step4: Determine the domain and range

The domain $D$ of the polynomial function $y=x^{4}-2x^{2}-2$ is all real numbers. Let $t=x^{2}$, then $y=t^{2}-2t - 2=(t - 1)^{2}-3$. Since $t\geq0$, the minimum value of $y$ occurs when $t = 1$ (i.e., $x=\pm1$) and $y=-3$. So the range $R=\{y|y\geq - 3\}$.

Answer:

Zeros: between $x=-2$ and $x=-1$, between $x = 1$ and $x = 2$; Relative maximum at $x = 0$; Relative minimum at $x=-1$ and $x = 1$; Domain: all real numbers; Range: $\{f(x)|f(x)\geq - 3\}$