Problem involving function
Given the function`f(x)=(x^4-3x^3-6x^2+7x-6) / (x^4+4x^3-5x)` , answer:
a) State the domain and the image of the function?
b) What are the asymptotes of this function?
c) Discuss the points of local maxima and minima of this function and on which intervals the function is growing and decreasing.
d) What are the inflection points and intervals of concavity of this function?
e) Sketch the graph of this function.
It fell in my assessment of the introduction to the calculation, only one person in the class hit the problem altogether. Anyone out there know how to solve this "bomb"?
Given `f(x)=(x^4-3x^3-6x^2+7x-6)/(x^4+4x^3-5x)` :
(1) The domain of a rational function is all real numbers except where the denominator is zero; solving `x^4+4x^3-5x=0` we get:
`x(x-1)(x^2+5x+5)=0==> x=0,1,1/2(-5+-sqrt(5))` . These are the only points not included in the domain.
The image is all real numbers.
(2) The vertical asymptotes occur when the denominator is zero and the numerator is nonzero. These are the points found in (1); the vertical asymptotes are `x=1/2(-5-sqrt(5)),x=1/2(-5+sqrt(5)),x=0,x=1`
Since the degree of the numerator is the same as the degree of the denominator,the horizontal asymptote occurs at the fraction of the leading coefficients: the horizontal asymptote is at y=1.
** You can also find the limit as `x->+-oo` ; divide numerator and denominator by `x^4` ; then as x tends to infinity all terms go to zero except the leading ones, so the limit is 1/1.
(3) To find the extrema and where the function is increasing or decreasing we need the first derivative. Using the quotient rule we have `f'(x)=(7x^6+12x^5-12x^4-2x^3+102x^2-30)/(x^2(x^3+4x^2-5)^2)`
Extrema occur when the first derivative is zero (or fails to exist -- in this case the only places f'(x) fails are not in the domain of f(x)). A fraction is zero if the numerator is zero and the denominator is not.
Solving `7x^6+12x^5-12x^4-2x^3+102x^2-30=0` with a graphing utility we find `f'(x)=0==> x~~-0.553115,x~~0.547979` . Using the first derivative test we find:
On `(-oo,1/2(-5-sqrt(5)))` that `f'(x)>0` so the function is increasing.
On `(-3.618,-1.382)` `f'(x)>0` so the function is increasing.
On `(-1.382,-0.553)` `f'(x)>0` while on `(-0.553,0)` `f'(x)<0` so the function has a local maximum at `x~~-0.553` .
On `(0,0.548)` `f'(x)<0` while on `(0.548,1)` `f'(x)>0` so the function has a relative minimum at `x~~0.548`
On `(1,oo)` `f'(x)>0` so the function is increasing.
(4) To find inflection points and concavity we need the second derivative. Again using the quotient rule we get:
Using a graphing utility we find the only real zero at `x~~-2.319` which is an inflection point.
Otherwise we find thesecond derivative to be positive on `(-oo,-3.61),(-2.32,-1.38),(0,1)` so the curve is concave up on those intervals and concave down on the rest.
(5) The x-intercepts are `x~~-2.09,x~~4.13` found by setting the numerator of f(x) equal to zero and using graphing technology to approximate the zeros.