Question

f left parenthesis x right parenthesis equals StartFraction 16 x squared Over x Superscript 4 Baseline plus 64 EndFractionf(x)=16x2 x4+64​(a) Is the pointleft parenthesis negative 2 StartRoot 2 EndRoot comma 1 right parenthesis−22,1on the graph of​ f?​(b) Ifx equals 2 commax=2,what is​ f(x)? What point is on the graph of​ f?​(c) If f left parenthesis x right parenthesis equals 1 commaf(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.

Answer 1

Answer:

a) Yes

b) (2,0.8)

c)$(2\sqrt2,1), (-2\sqrt2,1)$

d) $x \in (-\infty,\infty)$

e) (0,0)

f) (0,0)  

Step-by-step explanation:

We are given the following function in the question:

$f(x) = \displaystyle\frac{16x^2}{x^4 + 64}$

a) We have to check whether given point lies on the function or not.

$(-2\sqrt2,1)\\\\f(-2\sqrt2) = \displaystyle\frac{16(-2\sqrt2)^2}{(-2\sqrt2)^4 + 64} = \frac{128}{128} = 1$

b) Find value of f(x) at x = 2

$f(2) = \displaystyle\frac{16(2)^2}{(2)^4 + 64} =\frac{64}{80}= 0.8$

Thus, (2,0.8) lies on the graph of given function.

c) We have to find the value of x, when f(x) = 1

$1 = \displaystyle\frac{16x^2}{x^4 + 64}\\\\x^4 -16x^2 + 64 = 0\\(x^2-8)^2 = 0\\x^2 - 8 = 0\\x = \pm 2\sqrt2$

$(2\sqrt2,1), (-2\sqrt2,1)$ lies on he graph of function.

d) Domain is the collection of all values of x for which the function is defined.

$x \in (-\infty,\infty)$

e)  x-intercepts

This is the value of x such that the function is zero.

$0 = \displaystyle\frac{16x^2}{x^4 + 64}\\\\16x^2 = 0\\x = 0$

f) y-intercept

It is the value of function when x is zero.

$f(0) = \displaystyle\frac{16(0)^2}{(0)^4 + 64} = 0$

The function passes trough origin.