This problem is a good pre-exam example on functions, since it covers most of the aspects.
dom f(x)=dom(numerator)∪dom(denominator) - undefined points
Since both numerator and denominator are polynomials, dom is (-∞,∞). The denominator is positive for all real values of x, so there are no undefined (singular) points.
Therefore dom f(x) = (-∞,∞).
The function is continuous in its domain and the denominator is always positive, so there are no vertical asymptotes.
To find the horizontal asymptotes, if any, we take limits of the function as x->-∞(=3) and as x->+∞(=3). So there is a horizontal asymptote at y=3. To find from which side of y=3 the function approches the asymptote, we consider the term after the leading term in the numerator, -5x. Since -5x>0 when x->-∞ and -5x<0 when x->∞, we conclude that the function approaches the horizontal asymptote from 3+ as x->-∞ and from 3- as x->+∞.
x-intercepts, if any, will be the real roots of f(x)=0.
Since the function cannot be factorized, it is not obvious if there are x-intercepts. This can be confirmed after finding the local extrema following the calculation of critical points.
Since this is a function, there cannot be more than one y-intercept (ref. vertical line test).
The y-intercept equals f(0)=3.
f(x)≠f(-x), so it is not an even function, i.e. it is not symmetric about the y-axis.
f(x)≠-f(-x), so f(x) is not an odd function, i.e. it is not symmetrical with respect to the origin.
f(x) cannot be symmetrical about the x-axis, or else it will fail the vertical line test (i.e. not a function).
FIRST DERIVATIVE OF f(x)
The first derivative of f(x) can be calculated by the quotient rule and carefully simplifying the resulting expression.
f'(x) = 5(3x^4-1)/(x^4+1)²
If we factorize the numerator, we can easily find the zeroes of f'(x)=0:
from which the zeroes of f'(x) can be readily deduced as
We will denote these two values of x as
CRITICAL NUMBERS (or critical points) and END POINTS
If the function is defined on the interval [a,b], the points x=a and x=b are the end-points of the function. These are also points that are candidates of local or global extrema.
For the given function, the domain being (-∞,&infin), there are no end-points.
Critical points correspond to values of x for which f'(x)=0 or where f'(x) is undefined. Together with end-points, these are also the only points where local and global extrema can occur.
The only critical points for this function occur at x=c1 and x=c2. Since the horizontal asymptote is at y=3, the only places for a global maximum and global minimum can occur are at x=c1 and x=c2.
from f(c1)=5.84 (approx.) and f(c2)=0.15 (approx.), we conclude that the range of f(x) is [f(c2),f(c1)].
Since f(c1)>0 and f(c2)>0, and the horizontal asymptote y=3 >0, we conclude that f(x) has no real zeroes. SO THERE ARE NO X-INTERCEPTS.
Apply the first derivative test to confirm that (c1,f(c1)) is a local maximum, and (c2,f(c2)) is a local minimum.
Together with the horizontal asymptote, we conclude also that (c1,f(c1)) is a global maximum, and (c2,f(c2)) is a global minimum, since there are no end-points or other critical points.
SECOND DERIVATIVE OF f(x)
The second derivative is useful in
1. confirming the nature of the local extrema (local max. or minimum)
2. determining the concavity of the graph of the function.
3. determining the inflection points.
First, we need to find the second derivative of f(x).
This can be done again using the quotient rule and carefully simplifying the resulting expression, which becomes:
f"(c1)=-14.8 <0, therefore (c1,f(c1)) is a maximum.
f"(c2)=+14.8 >0 therefore (c2,f(c2)) is a minimum.
The above conclusions confirm the findings of the first derivative test.
Next, we determine the zero(es) of f"(x) which are readily found to be x=0, x=±(5/3)1/4 (=±1.14 approx.).
These will be denoted by
and they represent the points of inflection.
Since there are three inflection points, there are three changes of concavity.
For (-∞,i1], we evaluate f"(-2)=+1.4, therefore the graph is concave up. This also confirms the earlier observation that the graph approaches the horizontal asymptote from 3+.
For [i1,0], we evaluate f"(-1)=-5, meaning that the graph is concave down. Continuing similarly, we find [0,i2] is concave up, and [i2,∞] is concave down, again confirming that the graph approaches the horizontal asymptote from y=3-.
The graph of the function can be found at the following links: