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1. Domain:
Domain is the set of values of the independent variable for which the
function is defined. In this case the numerator is defined for all the
real numbers, but the division is not defined when the denominator is
zero, so this function is not defined when its denominator (x) is
equal to zero. Therefore, the domain of f(x) = (2x-1)/x is the set of
the real numbers except the zero.
2. Locate x and y intercepts:
A x-intercept is a value of x for which the function is zero, so we
must solve for f(x)=0 :
(2x-1)/x = 0 <==> 2x-1 = 0 <==> 2x = 1 <==> x = 1/2
1/2 is the only x-intercept.
An y-intercept is the value of the function when x is zero, but
remember that f(x) is not defined at such point, so you do not have
any y-intercept for this function.
3. Check for symmetry:
See the following definition:
"DEFINITION. A function f is even if the graph of f is symmetric with
respect to the y-axis. Algebraically, f is even if and only if f(-x) =
f(x) for all x in the domain of f.
A function f is odd if the graph of f is symmetric with respect to the
origin. Algebraically, f is odd if and only if f(-x) = -f(x) for all x
in the domain of f."
From "Visual Calculus - Even and Odd Functions":
http://archives.math.utk.edu/visual.calculus/0/functions.14/index.html
Since f(1/2) = 0 and f(-1/2) = 4, f is neither an even nor an odd function.
You can also see:
"Symmetry of graphs - Topics in precalculus":
http://www.themathpage.com/aPreCalc/symmetry.htm
4. Find all horizontal and vertical asymptotes:
x = c is a vertical asymptote if the following are all true:
1) f(c) is undefined
2) limit when x tends to c- of f(x) is +? or -?
3) limit when x tends to c+ of f(x) is +? or -?
The only choice here is x=0, the only point where f is undefined:
1) f(0) is undefined
2) limit when x tends to 0- of f(x) is +?
3) limit when x tends to c+ of f(x) is -?
We have a vertical asymptote at x=0.
If the limit when x tends to +? of f(x) = L OR the limit when x tends
to -? of f(x) = L, then the line y = L is a horizontal asymptote of f.
limit when x tends to +? of f(x) = 2
limit when x tends to -? of f(x) = 2
The line y = 2 is a horizontal asymptote of f.
See for references this page:
"1.1.5.1 Limits At Infinity And Infinite Limits":
http://www.geocities.com/pkving4math2tor1/1_lim_and_cont/1_01_05_01_lim_at_inf_and_inf_lim.htm
5. Locate all critical numbers:
Given a real number t. If f'(t) = 0 or f' is undefined at t, then t is
a critical number of f.
So start calculating f'(x):
f'(x) = [(2x-1)/x]' = (2 - 1/x)' = (2)' - (1/x)' = 1/x^2
Note that there is no number for which f'(x) = 0 and f'(x) is only
undefined at x=0, then x=0 as the only critical number of f (at a
later point you will see that for x<0 and x>0 the convexity of the
function changes).
6. Find all intervals on which f is increasing or decreasing:
Suppose that for a given interval is:
1) the first derivative of f is positive, then f is increasing;
2) the first derivative of f is negative, then f is decreasing.
f'(x) = 1/x^2 > zero in (-?,0) and (0,+?) (where f is defined), then f
is increasing in in all its domain.
7. Find all local max and min:
Since we have no critical numbers, except the zero when f is not
defined, we have no local max and min.
Note that because f is increasing in in all its domain, for a given
closed interval [a,b](a and b different to zero), the minimum is f(a)
and the maximum is f(b).
8. Find all intervals on which f is concave up and concave down:
Suppose that for a given interval is:
1) the second derivative of f is positive, then f is concave up;
2) the second derivative of f is negative, then f is concave down.
f''(x) = -2/x^3 , then:
f''(x) > 0 <==> x < 0, i.e. in (-?,0)
and
f''(x) < 0 <==> x > 0, i.e. in (0,+?)
So in (-?,0) f is concave up and in (0,+?) f is concave down.
9. Find all points of inflection:
A point of inflection is a point at which the graph changes from being
concave up to concave down, or vice versa. You can use the second
derivative to find them following this simple rule:
If (c, f(c)) is a point of inflection on the graph of f, then either
f''(c) = 0 or is undefined at x = c. Note that at a point of
inflection the second derivative must change its sign (this is the
test).
f''(x) = -2/x^3 , it is never zero and it is only undefined at x=0, so
zero is POSSIBLY a point of inflection. But f is not defined at zero
so it cannot be a point of inflection, then f has no inflections
points.
For references see the following pages:
"Graphing Using First and Second Derivatives":
A page with link to great solved examples!!!
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
"SparkNotes: Applications of the Derivative (AB)":
Start a navigation here to find explanations and examples related to this problem.
http://www.sparknotes.com/math/calcab/applicationsofthederivative/index.html
Search strategy:
I used the following keywords at Google.com:
function second derivative
function first derivative
points of inflection
intercepts x y
critical numbers
horizontal vertical asymptotes
domain function
I hope this helps you. Feel free to use the clarification feature if
you find something wrong, unclear or incomplete. I will be glad to
give you further assistance on this question if you need it.
Best regards,
livioflores-ga
function symmetry |
