Answer the questions about the following function. 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 point left parenthesis negative 2 StartRoot 2 EndRoot comma 1 right parenthesis−22,1 on the graph of f? (b) If x 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.
Accepted Solution
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Answer:Step-by-step explanation:(a) The function ... [tex]f(x)=\dfrac{16x^{2}}{x^{4}+64}[/tex]can be evaluated for x=-2√2 to get ... [tex]\displaystylef(-2\sqrt{2})=\frac{16(-2\sqrt{2})^{2}}{(-2\sqrt{2})^4+64}=\frac{128}{64+64}=1[/tex]The point (-2√2, 1) is on the graph of f(x).__(b) Likewise, we can evaluate for x=2: [tex]f(2)=\dfrac{16\cdot 2^2}{2^4+64}=\dfrac{64}{80}=0.8[/tex]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).