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 of f? (f) 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}. (b) if x = 1, f(x)=\frac{20}{101}. (simplify your answer.) using the information in the previous step, list the point(s) on the graph of f where x = 1. (simplify your answer. type an ordered pair. use a comma to separate answers as needed.)

Explanation:

Step1: Check if point is on graph for part (a)

Substitute $x = -\sqrt{10}$ into $f(x)=\frac{20x^{2}}{x^{4}+100}$.
$f(-\sqrt{10})=\frac{20\times(-\sqrt{10})^{2}}{(-\sqrt{10})^{4}+100}=\frac{20\times10}{100 + 100}=\frac{200}{200}=1$. So the point $(-\sqrt{10},1)$ is on the graph.

Step2: Find $f(x)$ when $x = 1$ for part (b)

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

Step3: Solve for $x$ when $f(x)=1$ for part (c)

Set $\frac{20x^{2}}{x^{4}+100}=1$. Then $20x^{2}=x^{4}+100$. Let $u = x^{2}$, so $u^{2}-20u + 100=0$. Factoring gives $(u - 10)^{2}=0$, so $u = 10$. Since $u=x^{2}$, $x=\pm\sqrt{10}$. The points on the graph are $(\sqrt{10},1),(-\sqrt{10},1)$.

Step4: Find domain for part (d)

The denominator $x^{4}+100$ is never zero for real - valued $x$ (because $x^{4}\geq0$ for all real $x$, so $x^{4}+100\geq100$). The domain of $f(x)$ is $(-\infty,\infty)$.

Step5: Find x - intercepts for part (e)

Set $f(x)=0$, so $\frac{20x^{2}}{x^{4}+100}=0$. Since the numerator $20x^{2}=0$ when $x = 0$, the x - intercept is $x = 0$.

Step6: Find y - intercept for part (f)

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

Answer:

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