Question

answer the questions about the following function. f(x) = \frac{20x^{2}}{x^{4}+100} (a) is the point (-\sqrt{10},1) on the graph of f? (b) if x = 1, what is f(x)? what point is on the graph of f? (c) if f(x)=1, what is x? what point(s) is/are on the graph of f? (d) what is the domain of f? (e) list the x - intercepts, if any, of the graph of f. (f) list the y - intercept, if there is one, of the graph of f. (a) choose the correct answer. a. yes, because f(-\sqrt{10}) = 1. b. yes, because f(1)=-\sqrt{10}. c. no, because f(-\sqrt{10})\
eq1. d. no, because f(1)\
eq-\sqrt{10}.

Explanation:

Step1: Substitute x = -√10 into f(x)

$f(-\sqrt{10})=\frac{20\times(-\sqrt{10})^{2}}{(-\sqrt{10})^{4}+100}=\frac{20\times10}{100 + 100}=\frac{200}{200}=1$

Step2: Analyze part (b)

When $x = 1$, $f(1)=\frac{20\times1^{2}}{1^{4}+100}=\frac{20}{101}$. The point on the graph is $(1,\frac{20}{101})$

Step3: Analyze part (c)

Set $f(x)=1$, so $\frac{20x^{2}}{x^{4}+100}=1$. Then $x^{4}-20x^{2}+100 = 0$. Let $u = x^{2}$, then $u^{2}-20u + 100=0$, which factors to $(u - 10)^{2}=0$. So $u=x^{2}=10$, and $x=\pm\sqrt{10}$. The points on the graph are $(\sqrt{10},1)$ and $(-\sqrt{10},1)$

Step4: Analyze part (d)

The denominator $x^{4}+100$ is always positive for all real - valued x since $x^{4}\geq0$ for all real x. So the domain of $f$ is $(-\infty,\infty)$

Step5: Analyze part (e)

Set $y = f(x)=0$, then $\frac{20x^{2}}{x^{4}+100}=0$. Since $x^{4}+100
eq0$, we solve $20x^{2}=0$, getting $x = 0$. The x - intercept is 0.

Step6: Analyze part (f)

Set $x = 0$, then $f(0)=\frac{20\times0^{2}}{0^{4}+100}=0$. The y - intercept is 0.

Answer:

(a) A. Yes, because $f(-\sqrt{10}) = 1$
(b) $f(1)=\frac{20}{101}$, point: $(1,\frac{20}{101})$
(c) $x=\pm\sqrt{10}$, points: $(\sqrt{10},1),(-\sqrt{10},1)$
(d) $(-\infty,\infty)$
(e) $x = 0$
(f) $y = 0$