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CALCULUS HELP NEEDED!!!?
g(x) = (2x-3.5)/(2x-12.25)^2
find the concavity, inflection points, x and y intercept , relative maximum and minimum ,increasing decreasing?
please explain your work I really want to learn this.
cheers!
2 Answers
- cidyahLv 77 years agoFavorite Answer
f(x) = (2x-3.5) /(2x-12.5)^2
x-intercept: set y=0
(2x-3.5) /(2x-12.5)^2 = 0
2x=3.5
x = 3.5/2 = 1.75
x-intercept : (1.75,0)
y-intercept: set x=0
y= -3.5/-12.5^2
y = -0.0224
y-intercept : (0, -0.0224)
f(x) = (2x-3.5) /(2x-12.5)^2
f(x) = (2x-3.5)(2x-12.5)^(-2)
f'(x) = (2) (2x-12.5)^(-2) + (2x-3.5) (-2)(2x-12.5)^(-3)(2)
f'(x) = 2 / (2x-12.5)^2 - 4 (2x-3.5) / (2x-12.5)^3 = 0
2(2x-12.5) / (2x-12.5)^3 - 4 (2x-3.5) / (2x-12.5)^3 = 0
[2(2x-12.5) - 4(2x-3.5)] / (2x-12.5)^3 = 0
[2(2x-12.5) - 4(2x-3.5)] = 0
-4x-25+14 =0
-4x = 11
x= -2.75 is the critical point
f'(x) = 2 / (2x-12.5)^2 - 4 (2x-3.5) / (2x-12.5)^3
f'(x) = 2(2x-12.5)^(-2) - 4(2x-3.5)(2x-12.5)^(-3)
f''(x) = (2)(-2)(2x-12.5)^(-3)(2) - 4(2)(2x-12.5)^(-3) - 4(2x-3.5)(-3)(2x-12.5)^(-4)(2)
f''(x) = -8 / (2x-12.5)^3 - 8 /(2x-12.5)^3 +24 (2x-3.5) /(2x-12.5)^4
f''(x) = -16 /(2x-3.5)^3 + 24 (2x-3.5) /(2x-12.5)^4
f''(-2.75) = 0.0199 > 0, so f has a relative minimum at x=-2.75
The minimum is f(-2.75) = -0.02777
Consider the intervals (-∞, -2.75),(-2.75, ∞)
choose any one point from each of the intervals.
if f'(x) < 0 , f(x) is decreasing on that interval
if f'(x) > 0 , f(x) is increasing on that interval
- Anonymous7 years ago
1. Read the Q
2. Solve the Q
3. ???
4. Profit
