F(x) = 2x^2-x-1
if f(x) = -1, what is x? what point(s) are on the graph
of f ?

Answers

Step-by-step explanation:

(a) The function ...

  f(x)=\dfrac{16x^{2}}{x^{4}+64}

can be evaluated for x=-2√2 to get ...

  \displaystylef(-2\sqrt{2})=\frac{16(-2\sqrt{2})^{2}}{(-2\sqrt{2})^4+64}=\frac{128}{64+64}=1

The point (-2√2, 1) is on the graph of f(x).

__

(b) Likewise, we can evaluate for x=2:

  f(2)=\dfrac{16\cdot 2^2}{2^4+64}=\dfrac{64}{80}=0.8

The point on the graph is (2, 0.8).

__

(c) From part (a), we know that f(-2√2) = 1. Since the function is even, this means that f(2√2) = 1. The graph is a maximum at those points, so there are no other values for which f(x) = 1.

The points (±2√2, 1) are on the graph.

__

(d) There are no values of x for which f(x) is undefined. The domain is all real numbers.

__

(e) The only x-intercept is at the origin, (0, 0). The x-axis is a horizontal asymptote.

__

(f) The only y-intercept is at the origin, (0, 0).


Answer the questions about the following function.  f left parenthesis x right parenthesis equals st

its B

Step-by-step explanation:

x = 0, x = 0.5 or 1/2 The points on the graph are ( -1/2 or -0.5,0 ) and ( 1,0 )

Step-by-step explanation:

2x^2-x-1 = -1

2x^2-x = 0

x(2x-1) = 0

x=0

x=0.5 or 1/2

x = 0, x = 0.5 or 1/2 The points on the graph are ( -1/2 or -0.5,0 ) and ( 1,0 )

Step-by-step explanation:

2x^2-x-1 = -1

2x^2-x = 0

x(2x-1) = 0

x=0

x=0.5 or 1/2



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