anonymous
  • anonymous
f(x) = (2) / (x^2-2x-3) Graph Domain Range X & Y intercepts Horizontal & Vertical Asymptote
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Domain: {x element R : x!=-1 and x!=3} Range: {f element R : f<=-1/2 or fɬ}
anonymous
  • anonymous
@timo86m
anonymous
  • anonymous
domain is all x values that make it work. In other words no division by zero find it thru this: (x^2-2x-3)=0 solve for x. and then say domain is every real number except x=(whatever you got as answer)

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anonymous
  • anonymous
https://www.google.com/search?q=(2)+%2F+(x%5E2-2x-3)&oq=(2)+%2F+(x%5E2-2x-3)&aqs=chrome..69i57.624j0&sourceid=chrome&ie=UTF-8 Here is your graph
anonymous
  • anonymous
Range is all y values that are allowable or that work. If you look at the graph of it. You can see it is all real numbers :)
anonymous
  • anonymous
So it's domain: all real numbers except x range: all real numbers except y
anonymous
  • anonymous
to find y intercept make x=0: just plug in 0 at x and evaluate (2) / (0^2-2*0-3)=-0.666 to find x intercept make y=0 i think there are non tho
anonymous
  • anonymous
Ohh i'm still confused
anonymous
  • anonymous
Find the derivative of the following via implicit differentiation: d/dx(f(x)) = d/dx(2/(x^2-2 x-3)) Rewrite the expression: d/dx(f(x)) = d/dx(f(x)): d/dx(f(x)) = d/dx(2/(-3-2 x+x^2)) The derivative of f(x) is f'(x): f'(x) = d/dx(2/(-3-2 x+x^2)) Factor out constants: f'(x) = 2 d/dx(1/(-3-2 x+x^2)) Using the chain rule, d/dx(1/(x^2-2 x-3)) = d/( du)1/u ( du)/( dx), where u = x^2-2 x-3 and ( d)/( du)(1/u) = -1/u^2: f'(x) = 2 -(d/dx(-3-2 x+x^2))/(x^2-2 x-3)^2 Simplify the expression: f'(x) = -(2 (d/dx(-3-2 x+x^2)))/(-3-2 x+x^2)^2 Differentiate the sum term by term and factor out constants: f'(x) = -(2 d/dx(-3)-2 d/dx(x)+d/dx(x^2))/(-3-2 x+x^2)^2 The derivative of -3 is zero: f'(x) = -(2 (-2 (d/dx(x))+d/dx(x^2)+0))/(-3-2 x+x^2)^2 Simplify the expression: f'(x) = -(2 (-2 (d/dx(x))+d/dx(x^2)))/(-3-2 x+x^2)^2 The derivative of x is 1: f'(x) = -(2 (d/dx(x^2)-1 2))/(-3-2 x+x^2)^2 Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x: f'(x) = -(2 (-2+2 x))/(-3-2 x+x^2)^2 Expand the left hand side: Answer: | | f'(x) = -(2 (-2+2 x))/(-3-2 x+x^2)^2
anonymous
  • anonymous
Maybe that will help =)
anonymous
  • anonymous
Lets do domain. Your graph/ function is a ratio. Takes the form c/a where c could be a function or number. In your case c is 2 where a is another function. Could be x could be x^2. in your case a is (x^2-2x-3) now lets consider a simple case. 1/x What is its domain? Domain is defined as all the x numbers that make the function 1/x work. Turns out all numbers except 0 work. Turns out that division by zero makes it undefined. now your case (2) / (x^2-2x-3) is same and there is not division by 0 x^2-2x-3=0 (x-3) (x+1) = 0 we factored so x= 3 and -1 Domain is all real numbers except 3 and -1
anonymous
  • anonymous
What's the range ?

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