Respuesta :
Answer:
a) Yes
b) (2,0.8)
c)[tex](2\sqrt2,1), (-2\sqrt2,1)[/tex]
d) [tex]x \in (-\infty,\infty)[/tex]
e) (0,0)
f) (0,0)
Step-by-step explanation:
We are given the following function in the question:
[tex]f(x) = \displaystyle\frac{16x^2}{x^4 + 64}[/tex]
a) We have to check whether given point lies on the function or not.
[tex](-2\sqrt2,1)\\\\f(-2\sqrt2) = \displaystyle\frac{16(-2\sqrt2)^2}{(-2\sqrt2)^4 + 64} = \frac{128}{128} = 1[/tex]
b) Find value of f(x) at x = 2
[tex]f(2) = \displaystyle\frac{16(2)^2}{(2)^4 + 64} =\frac{64}{80}= 0.8[/tex]
Thus, (2,0.8) lies on the graph of given function.
c) We have to find the value of x, when f(x) = 1
[tex]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[/tex]
[tex](2\sqrt2,1), (-2\sqrt2,1)[/tex] lies on he graph of function.
d) Domain is the collection of all values of x for which the function is defined.
[tex]x \in (-\infty,\infty)[/tex]
e) x-intercepts
This is the value of x such that the function is zero.
[tex]0 = \displaystyle\frac{16x^2}{x^4 + 64}\\\\16x^2 = 0\\x = 0[/tex]
f) y-intercept
It is the value of function when x is zero.
[tex]f(0) = \displaystyle\frac{16(0)^2}{(0)^4 + 64} = 0[/tex]
The function passes trough origin.
