I'm new here and I taking a course in calculus what is totally new to me, I having been work trough differentiation and now we got to do some graph sketching following 6 steps described in the book I have started this one but I'm not sure if I'm doing it correctly so all explanation and support in how to tackle and understand it is very much appreciated. So here we go
The Function f(x)= 6x-5 / 4−9x2
STEP 1 - Identify the domain of f
I came up with the domain is all R once 4−9x2 is never 0.
Is it correct?
STEP 2 - Wether f is even or odd
following my examples:
f(-x) = 6(-x)-5 / 4-9(-x)2
= -6x-5 / 4-9(x)2
Thus -f(x) the function is odd
STEP 3 - x and y intercepts
The x intercept is 6x-5=0
x = 0.833
The y intercept is f(0) = 6(0)-5 / 4-9(0)2
= -5 / 4
STEP 4 - The Intervals where x is positive or negative
f(x) = 6x-5 / 4-9x2
= 6x-5 / (2-3x)(2+3x)
x (-∞,-1.25) -1.25 (-1.25,0.833) 0.833 (0.833,∞) 6x-5 - - - - + 2-3x + - - - - 2+3x - + + + + f(x) + + + + -Thus f is positive on the Interval (-∞,-1.25) and negative on the interval(0.833,∞).
STEP 5 - The Intervals where f is increasing or decreasing and any stationary points, local maxima and local minima.
f ' (x) = (6)(-9x2+4) - (-18x)(6x-5) / (4-9x2)2
= 54x-66 / (4-9x2)2
x (-∞,-1.25) -1.25 (-1.25,0.833) 0.833 (0.833,∞) 54x2-66 - + - - + (4-9x2)2 + + + + + f ' (x) - + - - +
Thus f is decreasing on(-∞,-1.25)and increasing on(0.833,∞)also f has a stationary point at0.833which is a local minim with value f(0.833) = 0 and a stationary point at -1.25 which is a local maximum with value f(-1.25) = 1.242
STEP 6 - The asymptotic behaviour of f.
I can't understand how to discover the asymptote behaviours.
Well this is it does it all make sense to anyone? If it does what I have done is correct? If not please I really need to understand it and a step by step will really be appreciated.